WBCHSE Class 12 Maths Solutions For Some Special Integrals

WBCHSE Class 12 Maths Special Integrals

Some Special Integrals – Three Special Integrals

Theorem

(1) \(\int \frac{d x}{\left(a^2-x^2\right)}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\)

(2) \(\int \frac{d x}{\left(x^2-a^2\right)}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\)

(3) \(\int \frac{d x}{\left(x^2+a^2\right)}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C\)

Proof

(1) \(\int \frac{d x}{\left(a^2-x^2\right)}=\int \frac{d x}{(a+x)(a-x)}\)

= \(\int \frac{1}{2 a} \cdot\left\{\frac{(a-x)+(a+x)}{(a+x)(a-x)}\right\} d x=\frac{1}{2 a} \cdot\left[\int \frac{d x}{(a+x)}+\int \frac{d x}{(a-x)}\right]\)

= \(\frac{1}{2 a} \cdot[\log |a+x|-\log |a-x|]+C\)

= \(\frac{1}{2 a} \cdot \log \left|\frac{a+x}{a-x}\right|+C\)

∴ \(\int \frac{d x}{\left(a^2-x^2\right)}=\frac{1}{2 a} \cdot \log \left|\frac{a+x}{a-x}\right|+C\)

(2) \(\int \frac{d x}{\left(x^2-a^2\right)}=\int \frac{d x}{(x-a)(x+a)}\)

= \(\int \frac{1}{2 a} \cdot\left\{\frac{(x+a)-(x-a)}{(x-a)(x+a)}\right\} d x=\frac{1}{2 a} \cdot\left[\int \frac{d x}{(x-a)}-\int \frac{d x}{(x+a)}\right]\)

= \(\frac{1}{2 a} \cdot[\log |x-a|-\log |x+a|]+C=\frac{1}{2 a} \cdot \log \left|\frac{x-a}{x+a}\right|+C\)

∴ \(\int \frac{d x}{\left(x^2-a^2\right)}=\frac{1}{2 a} \cdot \log \left|\frac{x-a}{x+a}\right|+C\)

(3) \(\int \frac{d x}{\left(x^2+a^2\right)}=\frac{1}{a^2} \cdot \int \frac{d x}{\left(1+\frac{x^2}{a^2}\right)}\)

= \(\frac{1}{a^2} \cdot \int \frac{a d t}{\left(1+t^2\right)}\) [putting \(\frac{x}{a}\) = t and dx = a dt]

= \(\frac{1}{a} \tan ^{-1} t+C=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C .\)

∴ \(\int \frac{d x}{\left(x^2+a^2\right)}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C\)

Remark If we have an integral of the form \(\int \frac{d x}{\left(a x^2+b x+c\right)}\), then we put the denominator in the form [(x+α)2 ± β2] and then integrate.

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WBCHSE Class 12 Maths Special Integrals 

Solved Examples

Example 1 Evaluate:

(1) \(\int \frac{d x}{\left(1-4 x^2\right)}\)

(2) \(\int \frac{d x}{\left(32-2 x^2\right)}\)

(3) \(\int \frac{x^2}{\left(1-x^6\right)} d x\)

Solution We have

(1) \(\int \frac{d x}{\left(1-4 x^2\right)}=\frac{1}{4} \cdot \int \frac{d x}{\left(\frac{1}{4}-x^2\right)}\)

= \(\frac{1}{4} \cdot \int \frac{d x}{\left\{\left(\frac{1}{2}\right)^2-x^2\right\}}\)

= \(\frac{1}{4} \cdot \frac{1}{\left(2 \times \frac{1}{2}\right)} \cdot \log \left|\frac{\frac{1}{2}+x}{\frac{1}{2}-x}\right|+C\)

[∵ \(\int \frac{d x}{\left(a^2-x^2\right)}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\)]

\(\frac{1}{4} \log \left|\frac{1+2 x}{1-2 x}\right|+C\)

(2) \(\int \frac{d x}{\left(32-2 x^2\right)}=\frac{1}{2} \cdot \int \frac{d x}{\left(16-x^2\right)}\)

= \(\frac{1}{2} \cdot \int \frac{d x}{\left\{(4)^2-x^2\right\}}\)

= \(\frac{1}{2} \cdot \frac{1}{(2 \times 4)} \log \left|\frac{4+x}{4-x}\right|+C\)

[∵ \(\int \frac{d x}{\left(a^2-x^2\right)}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\)]

= \(\frac{1}{16} \log \left|\frac{4+x}{4-x}\right|+C\)

(3) Putting x3 = t and 3x2dx = dt, we get

\(\int \frac{x^2}{\left(1-x^6\right)} d x=\frac{1}{3} \cdot \int \frac{1}{\left(1-t^2\right)} d t\)

= \(\frac{1}{3} \cdot \frac{1}{(2 \times 1)} \cdot \log \left|\frac{1+t}{1-t}\right|+C\)

[∵ \(\int \frac{d x}{\left(a^2-x^2\right)}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\)]

= \(\frac{1}{6} \log \left|\frac{1+x^3}{1-x^3}\right|+C\)

WBCHSE Class 12 Maths Solutions For Some Special Integrals

WBBSE Class 12 Special Integrals Solutions

Example 2 Evaluate:

\(\int \frac{\sin x}{\left(1-4 \cos ^2 x\right)} d x\)

Solution

Putting cos x = t and -sin x dx = dt, we get

\(\int \frac{\sin x}{\left(1-4 \cos ^2 x\right)} d x=-\int \frac{d t}{\left(1-4 t^2\right)}\)

= \(-\frac{1}{4} \cdot \int \frac{d t}{\left(\frac{1}{4}-t^2\right)}=-\frac{1}{4} \cdot \int \frac{d t}{\left\{\left(\frac{1}{2}\right)^2-t^2\right\}}\)

= \(-\frac{1}{4} \log \left|\frac{1+2 t}{1-2 t}\right|+C=-\frac{1}{4} \log \left|\frac{1+2 \cos x}{1-2 \cos x}\right|+C\)

Example 3 Evaluate:

(1) \(\int \frac{d x}{\left(9 x^2-1\right)}\)

(2) \(\int \frac{x}{\left(x^4-9\right)} d x\)

(3) \(\int \frac{x^2}{\left(x^2-9\right)} d x\)

Solution We have

(1) \(\int \frac{d x}{\left(9 x^2-1\right)}=\frac{1}{9} \cdot \int \frac{d x}{\left(x^2-\frac{1}{9}\right)}\)

= \(\frac{1}{9} \cdot \int \frac{d x}{\left\{x^2-\left(\frac{1}{3}\right)^2\right\}}\)

= \(\frac{1}{9} \cdot \frac{1}{\left(2 \times \frac{1}{3}\right)} \log \left|\frac{x-\frac{1}{3}}{x+\frac{1}{3}}\right|+C\)

[∵ \(\int \frac{d x}{\left(x^2-a^2\right)}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\)]

(2) Putting x2 = t and 2x dx = dt, we get

\(\int \frac{x}{\left(x^4-9\right)} d x=\frac{1}{2} \int \frac{d t}{\left(t^2-9\right)}=\frac{1}{2} \cdot \int \frac{d t}{\left\{t^2-(3)^2\right\}}\)

= \(\frac{1}{2} \cdot \frac{1}{(2 \times 3)} \log \left|\frac{t-3}{t+3}\right|+C\)

= \(\frac{1}{12} \log \left|\frac{x^2-3}{x^2+3}\right|+C\)

(3) \(\int \frac{x^2}{\left(x^2-9\right)} d x=\int\left\{1+\frac{9}{x^2-9}\right\} d x\)

= \(\int d x+9 \int \frac{d x}{\left[x^2-(3)^2\right]}\)

= \(x+9 \cdot\left[\frac{1}{(2 \times 3)} \log \left|\frac{x-3}{x+3}\right|\right]+C\)

= \(x+\frac{3}{2} \log \left|\frac{x-3}{x+3}\right|+C\).

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NEET Foundation Class 12 Physics NEET Physics

Example 4 Evaluate \(\int \frac{d x}{\left(4+25 x^2\right)} .\)

Solution We have

\(\int \frac{d x}{\left(4+25 x^2\right)}=\frac{1}{25} \cdot \int \frac{d x}{\left(\frac{4}{25}+x^2\right)}\)

= \(\frac{1}{25} \cdot \int \frac{d x}{\left\{(2 / 5)^2+x^2\right\}}\)

[∵ \(\int \frac{d x}{\left(a^2+x^2\right)}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+\mathrm{C}\)]

= \(\frac{1}{10} \tan ^{-1}\left(\frac{5 x}{2}\right)+C\).

WBCHSE Class 12 Maths Special Integrals 

Example 5 Evaluate \(\int \frac{3 x}{\left(1+2 x^4\right)} d x\)

Solution

\(\int \frac{3 x}{\left(1+2 x^4\right)} d x\)

Putting x2 = t and 2x dx = dt, we get

\(\int \frac{3 x}{\left(1+2 x^4\right)} d x=\frac{3}{2} \cdot \int \frac{d t}{\left(1+2 t^2\right)}\)

= \(\frac{3}{4} \cdot \int \frac{d t}{\left(\frac{1}{2}+t^2\right)}=\frac{3}{4} \cdot \int \frac{d t}{\left\{\left(\frac{1}{\sqrt{2}}\right)^2+t^2\right\}}\)

= \(\frac{3}{4} \cdot \frac{1}{(1 / \sqrt{2})} \tan ^{-1} \frac{t}{(1 / \sqrt{2})}+C\)

= \(\frac{3}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} t)+C=\frac{3}{2 \sqrt{2}} \tan ^{-1}\left(\sqrt{2} x^2\right)+C\).

\(\int \frac{3 x}{\left(1+2 x^4\right)} d x\) = \(\frac{3}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} t)+C=\frac{3}{2 \sqrt{2}} \tan ^{-1}\left(\sqrt{2} x^2\right)+C\).

Example 6 Evaluate \(\int \frac{d x}{\left(1+5 \sin ^2 x\right)}\).

Solution

\(\int \frac{d x}{\left(1+5 \sin ^2 x\right)}\)

On dividing num. and denom. by cos2x, we get

\(\int \frac{d x}{\left(1+5 \sin ^2 x\right)}=\int \frac{\left(1 / \cos ^2 x\right)}{\left(\frac{1}{\cos ^2 x}+5 \cdot \frac{\sin ^2 x}{\cos ^2 x}\right)} d x\)

= \(\int \frac{\sec ^2 x}{\left(\sec ^2 x+5 \tan ^2 x\right)} d x=\int \frac{\sec ^2 x}{\left\{\left(1+\tan ^2 x\right)+5 \tan ^2 x\right\}} d x\)

= \(\int \frac{\sec ^2 x}{\left(1+6 \tan ^2 x\right)} d x=\int \frac{d t}{\left(1+6 t^2\right)}\), where tan x = t

= \(\frac{1}{6} \int \frac{d t}{\left(\frac{1}{6}+t^2\right)}=\frac{1}{6} \cdot \int \frac{d t}{\left\{\left(\frac{1}{\sqrt{6}}\right)^2+t^2\right\}}\)

= \(\frac{1}{6} \cdot \frac{1}{(1 / \sqrt{6})} \tan ^{-1} \frac{t}{(1 / \sqrt{6})}+C=\frac{1}{\sqrt{6}} \tan ^{-1}(\sqrt{6} t)+C\)

= \(\frac{1}{\sqrt{6}} \tan ^{-1}(\sqrt{6} \tan x)+C \text {. }\)

\(\int \frac{d x}{\left(1+5 \sin ^2 x\right)}\) = \(\frac{1}{\sqrt{6}} \tan ^{-1}(\sqrt{6} \tan x)+C \text {. }\)

Step-by-Step Solutions to Special Integral Problems

Example 7 Evaluate \(\int \frac{d x}{\left(2+\sin ^2 x\right)}\)

Solution

\(\int \frac{d x}{\left(2+\sin ^2 x\right)}\)

On dividing num. and denom. by cos2x, we get

\(\int \frac{d x}{\left(2+\sin ^2 x\right)}=\int \frac{\sec ^2 x}{\left(2 \sec ^2 x+\tan ^2 x\right)} d x=\int \frac{\sec ^2 x}{\left\{2\left(1+\tan ^2 x\right)+\tan ^2 x\right\}} d x\)

= \(\int \frac{\sec ^2 x}{2+3 \tan ^2 x} d x=\int \frac{d t}{\left(2+3 t^2\right)}\), where tan x = t

= \(\frac{1}{3} \cdot \int \frac{d t}{\left(t^2+\frac{2}{3}\right)}=\frac{1}{3} \cdot \int \frac{d t}{\left(\sqrt{\frac{2}{3}}\right)^2+t^2}\)

= \(\frac{1}{3} \cdot \frac{1}{\sqrt{\frac{2}{3}}} \cdot \tan ^{-1} \frac{t}{\left(\sqrt{\frac{2}{3}}\right)}+C=\frac{1}{\sqrt{6}} \tan ^{-1} \frac{\sqrt{3} t}{\sqrt{2}}+C\)

= \(\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{2}}\right)+C\).

\(\int \frac{d x}{\left(2+\sin ^2 x\right)}\) = \(\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{2}}\right)+C\).

Example 8 Evaluate \(\int \frac{d x}{\left(x^2+6 x+13\right)}\).

Solution We have

\(\int \frac{d x}{\left(x^2+6 x+13\right)}=\int \frac{d x}{\left\{\left(x^2+6 x+9\right)+4\right\}}\)

= \(\int \frac{d x}{\left\{(x+3)^2+2^2\right\}}=\int \frac{d t}{\left(t^2+2^2\right)}\), where (x+3) = t

= \(\frac{1}{2} \tan ^{-1} \frac{t}{2}+C\)

= \(\frac{1}{2} \tan ^{-1} \frac{1}{2}(x+3)+C .\)

\(\int \frac{d x}{\left(x^2+6 x+13\right)}\) = \(\frac{1}{2} \tan ^{-1} \frac{1}{2}(x+3)+C .\)

Example 9 Evaluate \(\int \frac{d x}{\left(x^2+8 x+20\right)}\)

Solution We have

\(\int \frac{d x}{\left(x^2+8 x+20\right)}=\int \frac{d x}{\left\{(x+4)^2+2^2\right\}}=\int \frac{d t}{\left(t^2+2^2\right)}\), where (x+4) = t-3

= \(\frac{1}{2} \tan ^{-1} \frac{t}{2}+C\)

= \(\frac{1}{2} \tan ^{-1} \frac{1}{2}(x+4)+C\).

\(\int \frac{d x}{\left(x^2+8 x+20\right)}\) = \(\frac{1}{2} \tan ^{-1} \frac{1}{2}(x+4)+C\).

Example 10 Evaluate \(\int \frac{d x}{\left(9 x^2-12 x+8\right)}\)

Solution We have

\(\left(9 x^2-12 x+8\right)=9\left(x^2-\frac{4}{3} x+\frac{8}{9}\right)\)

= \(9\left\{\left(x^2-\frac{4}{3} x+\frac{4}{9}\right)-\frac{4}{9}+\frac{8}{9}\right\}=9\left\{\left(x-\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^2\right\}\)

∴ \(\int \frac{d x}{\left(9 x^2-12 x+8\right)}=\frac{1}{9} \cdot \int \frac{d x}{\left\{\left(x-\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^2\right\}}\)

= \(\frac{1}{9} \cdot \frac{1}{(2 / 3)} \tan ^{-1}\left\{\frac{\left(x-\frac{2}{3}\right)}{(2 / 3)}\right\}+C=\frac{1}{6} \tan ^{-1}\left(\frac{3 x-2}{2}\right)+C\).

\(\int \frac{d x}{\left(9 x^2-12 x+8\right)}\) = \(\frac{1}{9} \cdot \frac{1}{(2 / 3)} \tan ^{-1}\left\{\frac{\left(x-\frac{2}{3}\right)}{(2 / 3)}\right\}+C=\frac{1}{6} \tan ^{-1}\left(\frac{3 x-2}{2}\right)+C\).

WBCHSE Class 12 Maths Special Integrals 

Example 11 Evaluate \(\int \frac{x}{\left(x^4-x^2+1\right)} d x\).

Solution

\(\int \frac{x}{\left(x^4-x^2+1\right)} d x\)

Putting x2 = t and 2x dx = dt, we get

\(\int \frac{x}{\left(x^4-x^2+1\right)} d x=\frac{1}{2} \cdot \int \frac{d t}{\left(t^2-t+1\right)}=\frac{1}{2} \cdot \int \frac{d t}{\left\{\left(t-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right\}}\)

= \(\frac{1}{2} \cdot \frac{1}{\left(\frac{\sqrt{3}}{2}\right)} \tan ^{-1} \frac{\left(t-\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}+C\)

= \(\frac{1}{\sqrt{3}} \cdot \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^2-1}{\sqrt{3}}\right)+C\)

\(\int \frac{x}{\left(x^4-x^2+1\right)} d x\). = \(\frac{1}{\sqrt{3}} \cdot \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^2-1}{\sqrt{3}}\right)+C\)

Example 12 Evaluate \(\int \frac{d x}{\left(2 x^2+x-1\right)}\).

Solution We have

\(\int \frac{d x}{\left(2 x^2+x-1\right)}=\frac{1}{2} \cdot \int \frac{d x}{\left(x^2+\frac{1}{2} x-\frac{1}{2}\right)}\)

= \(\frac{1}{2} \int \frac{d x}{\left[\left\{x^2+\frac{1}{2} x+\left(\frac{1}{4}\right)^2\right\}-\frac{1}{16}-\frac{1}{2}\right]}=\frac{1}{2} \int \frac{d x}{\left[\left(x+\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2\right]}\)

= \(\frac{1}{2} \cdot \frac{1}{2 \cdot \frac{3}{4}} \log \left|\frac{\left(x+\frac{1}{4}\right)-\frac{3}{4}}{\left(x+\frac{1}{4}\right)+\frac{3}{4}}\right|+C=\frac{1}{3} \log \left|\frac{2 x-1}{2(x+1)}\right|+C\)

Common Formulas for Special Integrals

Example 13 Evaluate \(\int \frac{d x}{\left(3 x^2+13 x-10\right)}\).

Solution We have

\(\left(3 x^2+13 x-10\right)=3\left(x^2+\frac{13}{3} x-\frac{10}{3}\right)\)

= \(3\left\{\left(x+\frac{13}{6}\right)^2-\frac{169}{36}-\frac{10}{3}\right\}=3\left\{\left(x+\frac{13}{6}\right)^2-\frac{289}{36}\right\}\)

= \(3\left\{\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\right\}.\)

∴ \(\int \frac{d x}{\left(3 x^2+13 x-10\right)}=\int \frac{d x}{3\left\{\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\right\}}\)

= \(\frac{1}{3} \int \frac{d t}{\left\{t^2-\left(\frac{17}{6}\right)^2\right\}}\), where [x+\(\frac{13}{6}\)] = t [/latex]

= \(\frac{1}{3} \cdot \frac{1}{\left(2 \times \frac{17}{6}\right)} \log \left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+C\)

[∵ \(\int \frac{d x}{\left(x^2-a^2\right)}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\)]

= \(\frac{1}{17} \log \left|\frac{6 t-17}{6 t+17}\right|+C\)

= \(\frac{1}{17} \log \left|\frac{6\left(x+\frac{13}{6}\right)-17}{6\left(x+\frac{13}{6}\right)+17}\right|=\frac{1}{17} \log \left|\frac{6 x-4}{6 x+30}\right|+C\)

= \(\frac{1}{17} \log \left|\frac{3 x-2}{3 x+15}\right|+C=\frac{1}{17} \log \left|\frac{(3 x-2)}{3(x+5)}\right|+C\)

= \(\frac{1}{17}\left\{\log \frac{1}{3}+\log \left|\frac{3 x-2}{x+5}\right|\right\}+C\)

= \(\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+k\),

where \(\frac{1}{17} \log \frac{1}{3}+C\) = k = constant.

WBCHSE Class 12 Maths Special Integrals 

Example 14 Evaluate \(\int \frac{d x}{\left(1+x-x^2\right)} .\)

Solution We have

\(\int \frac{d x}{\left(1+x-x^2\right)}=-\int \frac{d x}{\left(x^2-x-1\right)}\)

= \(-\int \frac{d x}{\left\{\left(x^2-x+\frac{1}{4}\right)-\frac{5}{4}\right\}}=-\int \frac{d x}{\left\{\left(x-\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2\right\}}\)

= \(\int \frac{d x}{\left\{\left(\frac{\sqrt{5}}{2}\right)^2-\left(x-\frac{1}{2}\right)^2\right\}}=\int \frac{d x}{\left\{\left(\frac{\sqrt{5}}{2}\right)^2-u^2\right\}}\), where [x – \(\frac{1}{2}\)] = u

= \(\frac{1}{\left(2 \times \frac{\sqrt{5}}{2}\right)} \cdot \log \left|\frac{\frac{\sqrt{5}}{2}+u}{\frac{\sqrt{5}}{2}-u}\right|+C\)

= \(\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{5}+2 u}{\sqrt{5}-2 u}\right|+C=\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{5}+2\left(x-\frac{1}{2}\right)}{\sqrt{5}-2\left(x-\frac{1}{2}\right)}\right|+C\)

= \(\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{5}+2 x-1}{\sqrt{5}-2 x+1}\right|+C=\frac{1}{\sqrt{5}} \log \left|\frac{(\sqrt{5}-1)+2 x}{(\sqrt{5}+1)-2 x}\right|+C .\)

Example 15 Evaluate \(\int \frac{d x}{\left(5-8 x-x^2\right)}\)

Solution We have

\(\int \frac{d x}{\left(5-8 x-x^2\right)}=-\int \frac{d x}{\left(x^2+8 x-5\right)}\)

= \(-\int \frac{d x}{\left\{\left(x^2+8 x+16\right)-21\right\}}=-\int \frac{d x}{\left\{(x+4)^2-(\sqrt{21})^2\right\}}\)

= \(\int \frac{d x}{\left\{(\sqrt{21})^2-(x+4)^2\right\}}=\int \frac{d t}{\left\{(\sqrt{21})^2-t^2\right\}}\), Where (x+4) = t

= \(\frac{1}{2 \sqrt{21}} \cdot \log \left|\frac{\sqrt{21}+t}{\sqrt{21}-t}\right|+C\)

= \(\frac{1}{2 \sqrt{21}} \cdot \log \left|\frac{\sqrt{21}+4+x}{\sqrt{21}-4-x}\right|+C\).

Example 16 Evaluate \(\int \frac{d x}{\left(1-6 x-9 x^2\right)} .\)

Solution We have

\(\int \frac{d x}{\left(1-6 x-9 x^2\right)}=-\int \frac{d x}{\left(9 x^2+6 x-1\right)}=-\frac{1}{9} \int \frac{d x}{\left(x^2+\frac{2}{3} x-\frac{1}{9}\right)}\)

= \(-\frac{1}{9} \cdot \int \frac{d x}{\left\{\left(x^2+\frac{2}{3} x+\frac{1}{9}\right)-\frac{2}{9}\right\}}\)

= \(-\frac{1}{9} \cdot \int \frac{d x}{\left\{\left(x+\frac{1}{3}\right)^2-\left(\frac{\sqrt{2}}{3}\right)^2\right\}}=\frac{1}{9} \cdot \int \frac{d x}{\left\{\left(\frac{\sqrt{2}}{3}\right)^2-\left(x+\frac{1}{3}\right)^2\right\}}\)

= \(\frac{1}{9} \cdot \int \frac{d x}{\left\{\left(\frac{\sqrt{2}}{3}\right)^2-t^2\right\}}\), where (x + \(\frac{1}{3}\)) = t

= \(\frac{1}{9} \cdot \frac{1}{2 \cdot \frac{\sqrt{2}}{3}} \log \left|\frac{\frac{\sqrt{2}}{3}+t}{\frac{\sqrt{2}}{3}-t}\right|+C=\frac{1}{6 \sqrt{2}} \log \left|\frac{\sqrt{2}+3 t}{\sqrt{2}-3 t}\right|+C\)

= \(\frac{1}{6 \sqrt{2}} \log \left|\frac{\sqrt{2}+3\left(x+\frac{1}{3}\right)}{\sqrt{2}-3\left(x+\frac{1}{3}\right)}\right|+C\)

= \(\frac{1}{6 \sqrt{2}} \log \left|\frac{\sqrt{2}+1+3 x}{\sqrt{2}-1-3 x}\right|+C.\)

Class 12 Integration Special Integrals 

Example 17 Evaluate \(\int \frac{\cos x}{\left(\sin ^2 x+4 \sin x+5\right)} d x\)

Solution

\(\int \frac{\cos x}{\left(\sin ^2 x+4 \sin x+5\right)} d x\)

Putting sin x = t and cos x dx = dt, we get

\(\int \frac{\cos x}{\left(\sin ^2 x+4 \sin x+5\right)} d x=\int \frac{d t}{\left(t^2+4 t+5\right)}=\int \frac{d t}{\left\{\left(t^2+4 t+4\right)+1\right\}}\)

= \(\int \frac{d t}{\left\{(t+2)^2+1^2\right\}}=\int \frac{d u}{\left(u^2+1\right)}\), where u = (t+2)

= tan-1u + C = tan-1(t+2) + C

= tan-1(sin x + 2) + C.

Example 18 \(\int \frac{e^x}{\left(e^{2 x}+6 e^x+5\right)} d x\).

Solution

\(\int \frac{e^x}{\left(e^{2 x}+6 e^x+5\right)} d x\)

Putting ex = t and exdx = dt, we get

\(\int \frac{e^x}{e^{2 x}+6 e^x+5} d x=\int \frac{d t}{\left(t^2+6 t+5\right)}=\int \frac{d t}{\left\{\left(t^2+6 t+9\right)-4\right\}}\)

= \(\int \frac{d t}{\left\{(t+3)^2-2^2\right\}}=\int \frac{d u}{\left(u^2-2^2\right)}\), where (t+3) = u

= \(\frac{1}{(2 \times 2)} \log \left|\frac{u-2}{u+2}\right|+C=\frac{1}{4} \log \left|\frac{t+3-2}{t+3+2}\right|+C\)

= \(\frac{1}{4} \log \left|\frac{t+1}{t+5}\right|+C=\frac{1}{4} \log \left|\frac{e^x+1}{e^x+5}\right|+C .\)

Examples of Special Integrals with Solutions

Integrals of the form \(\int \frac{(p x+q)}{\left(a x^2+b x+c\right)} d x\)

Method Let (px + q) = A. \(\frac{d}{d x}\left(a x^2+b x+c\right)\) + B

Find A and B.

Now, the integrand so obtained can be integrated easily.

Class 12 Integration Special Integrals 

Example 19 Evaluate \(\int \frac{x}{\left(x^2+x+1\right)} d x\)

Solution

Let x = A . \(\frac{d}{d x}\left(x^2+x+1\right)\) + B. Then,

x = A(2x+1) + B.

Comparing coefficients of like powers of x, we get

(2A = 1 and A + B = 0) ⇒ [A=\(\frac{1}{2}\) and B=\(\frac{-1}{2}\)].

∴ \(\int \frac{x}{\left(x^2+x+1\right)} d x=\int \frac{\frac{1}{2} \cdot(2 x+1)-\frac{1}{2}}{\left(x^2+x+1\right)} d x\)

= \(\frac{1}{2} \int \frac{(2 x+1)}{\left(x^2+x+1\right)} d x-\frac{1}{2} \int \frac{d x}{\left(x^2+x+1\right)}\)

= \(\frac{1}{2} \log \left|x^2+x+1\right|-\frac{1}{2} \int \frac{d x}{\left\{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right\}}\)

= \(\frac{1}{2} \log \left|x^2+x+1\right|-\frac{1}{2} \cdot \frac{1}{\left(\frac{\sqrt{3}}{2}\right)} \tan ^{-1}\left\{\frac{\left(x+\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}\right\}+C\)

= \(\frac{1}{2} \log \left|x^2+x+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C .\)

Example 20 Evaluate \(\int \frac{(3 x+1)}{\left(2 x^2-2 x+3\right)} d x\)

Solution

\(\int \frac{(3 x+1)}{\left(2 x^2-2 x+3\right)} d x\)

Let (3x+1) = A . \(\frac{d}{d x}\left(2 x^2-2 x+3\right)\) + B. Then,

(3x+1) = A(4x-2) + B

Comparing coefficients of like powers of x, we get

(4A = 3 and B – 2A = 1) ⇒ [A=\(\frac{3}{4}\) and B=\(\frac{5}{2}\)].

∴ \(\int \frac{(3 x+1)}{\left(2 x^2-2 x+3\right)} d x=\int \frac{A \cdot(4 x-2)+B}{\left(2 x^2-2 x+3\right)}\)

= \(\int \frac{\frac{3}{4} \cdot(4 x-2)+\frac{5}{2}}{\left(2 x^2-2 x+3\right)} d x=\frac{3}{4} \cdot \int \frac{(4 x-2)}{\left(2 x^2-2 x+3\right)} d x+\frac{5}{2} \int \frac{d x}{2\left(x^2-x+\frac{3}{2}\right)}\)

= \(\frac{3}{4} \log \left|2 x^2-2 x+3\right|+\frac{5}{4} \cdot \int \frac{d x}{\left\{\left(x^2-x+\frac{1}{4}\right)+\frac{5}{4}\right\}}\)

= \(\frac{3}{4} \log \left|2 x^2-2 x+3\right|+\frac{5}{4} \cdot \int \frac{d x}{\left\{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{5}}{2}\right)^2\right\}}\)

= \(\frac{3}{4} \log \left|2 x^2-2 x+3\right|+\frac{5}{4} \cdot \frac{1}{\left(\frac{\sqrt{5}}{2}\right)} \tan ^{-1}\left\{\frac{\left(x-\frac{1}{2}\right)}{\left(\frac{\sqrt{5}}{2}\right)}\right\}+C\)

= \(\frac{3}{4} \log \left|2 x^2-2 x+3\right|+\frac{\sqrt{5}}{2} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)+C \text {. }\)

Class 12 Integration Special Integrals 

Example 21 Evaluate \(\int \frac{(2 x+1)}{\left(4-3 x-x^2\right)} d x\)

Solution

\(\int \frac{(2 x+1)}{\left(4-3 x-x^2\right)} d x\)

Let (2x+1) = A . \(\frac{d}{d x}\left(4-3 x-x^2\right)\) + B.

Then, (2x+1) = A(-3-2x) + B …(1)

Comparing coefficients of like terms, we get

(-2A = 2 and -3A + B = 1) ⇒ (A = -1, B = -2).

∴ \(\int \frac{(2 x+1)}{\left(4-3 x-x^2\right)} d x=\int\left\{\frac{(-1) \cdot(-3-2 x)-2}{\left(4-3 x-x^2\right)}\right\} d x\)

= \(-\int \frac{(-3-2 x)}{\left(4-3 x-x^2\right)} d x-2 \int \frac{d x}{\left(4-3 x-x^2\right)}\)

= \(-\log \left|4-3 x-x^2\right|+2 \int \frac{d x}{\left(x^2+3 x-4\right)}\)

= \(-\log \left|4-3 x-x^2\right|+2 \int \frac{d x}{\left(x+\frac{3}{2}\right)^2-\left(4+\frac{9}{4}\right)}\)

= \(-\log \left|4-3 x-x^2\right|+2 \int \frac{d x}{\left\{\left(x+\frac{3}{2}\right)^2-\left(\frac{5}{2}\right)^2\right\}}\)

= \(-\log \left|4-3 x-x^2\right|+\frac{2}{\left(2 \times \frac{5}{2}\right)} \log \left|\frac{\left(x+\frac{3}{2}\right)-\frac{5}{2}}{\left(x+\frac{3}{2}\right)+\frac{5}{2}}\right|+C\)

= \(-\log \left|4-3 x-x^2\right|+\frac{2}{5} \log \left|\frac{x-1}{x+4}\right|+C\).

Example 22 Evaluate \(\int\left(\frac{x^2+5 x+3}{x^2+3 x+2}\right) d x\).

Solution We have

\(\frac{\left(x^2+5 x+3\right)}{\left(x^2+3 x+2\right)}=\left\{1+\frac{(2 x+1)}{x^2+3 x+2}\right\}\)

⇒ \(\int \frac{\left(x^2+5 x+3\right)}{\left(x^2+3 x+2\right)} d x=\int d x+\int \frac{(2 x+1)}{\left(x^2+3 x+2\right)} d x\) …(1)

Let (2x+1) = A . \(\frac{d}{d x}\left(x^2+3 x+2\right)+B.\) Then,

(2x+1) = A(2x+3) + B …(2)

Comparing coefficients of like powers of x, we get

(2A = 2 and 3A + B = 1) ⇒ (A=1 and B=-2).

∴ (2x+1) = (2x+3) – 2.

∴ \(I=x+\int \frac{(2 x+1)}{\left(x^2+3 x+2\right)} d x\)

= \(x+\int \frac{\{(2 x+3)-2\}}{\left(x^2+3 x+2\right)} d x=x+\int \frac{(2 x+3)}{\left(x^2+3 x+2\right)} d x-2 \int \frac{d x}{\left(x^2+3 x+2\right)}\)

= \(x+\log \left|x^2+3 x+2\right|-2 \int \frac{d x}{\left\{\left(x^2+3 x+\frac{9}{4}\right)+\left(2-\frac{9}{4}\right)\right\}}\)

= \(x+\log \left|x^2+3 x+2\right|-2 \int \frac{d x}{\left\{\left(x+\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2\right\}}\)

= \(x+\log \left|x^2+3 x+2\right|-2 \cdot \frac{1}{\left(2 \times \frac{1}{2}\right)} \log \left|\frac{x+\frac{3}{2}-\frac{1}{2}}{x+\frac{3}{2}+\frac{1}{2}}\right|+C\)

= \(x+\log \left|x^2+3 x+2\right|-2 \log \left|\frac{x+1}{x+2}\right|+C.\)

Applications of Special Integrals in Mathematics

Integrals of the form \(\int \frac{d x}{a+b \cos ^2 x}, \int \frac{d x}{a+b \sin ^2 x}\)

and \(\int \frac{d x}{a \cos ^2 x+b \sin x \cos x+c \sin ^2 x} .\)

Method In each such an integral, we divide the numerator and denominator by cos2x and put tan x = t, sec2x dx = dt and then integrate.

Example 23 Evaluate \(\int \frac{d x}{a^2 \sin ^2 x+b^2 \cos ^2 x}\)

Solution

\(\int \frac{d x}{a^2 \sin ^2 x+b^2 \cos ^2 x}\)

Dividing the numerator and the denominator of the given integrand by cos2x, we get

\(\int \frac{d x}{\left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)}=\int \frac{\sec ^2 x}{a^2 \tan ^2 x+b^2} d x\)

= \(\int \frac{d t}{\left(a^2 t^2+b^2\right)}\) [putting tan x = t]

= \(\frac{1}{a^2} \int \frac{d t}{\left[t^2+\left(\frac{b}{a}\right)^2\right]}=\frac{1}{a^2} \cdot \frac{1}{\left(\frac{b}{a}\right)} \tan ^{-1} \frac{t}{\left(\frac{b}{a}\right)}+C\)

= \(\frac{1}{a b} \tan ^{-1}\left(\frac{a t}{b}\right)+C=\frac{1}{a b} \tan ^{-1}\left(\frac{a \tan x}{b}\right)+C\).

Example 24 Evaluate:

(1) \(\int \frac{d x}{\left(1+3 \sin ^2 x\right)}\)

(2) \(\int \frac{d x}{\left(3+2 \cos ^2 x\right)}\)

Solution

(1) Dividing the numerator and denominator by cos2x, we get

\(\int \frac{d x}{\left(1+3 \sin ^2 x\right)}=\int \frac{\sec ^2 x}{\sec ^2 x+3 \tan ^2 x} d x=\int \frac{\sec ^2 x}{\left(1+4 \tan ^2 x\right)} d x\)

= \(\int \frac{d t}{\left(1+4 t^2\right)}\) [putting tan x = t]

= \(\frac{1}{4} \int \frac{d t}{\left[t^2+\left(\frac{1}{2}\right)^2\right]}\)=\(\frac{1}{4} \cdot \frac{1}{(1 / 2)} \tan ^{-1} \frac{t}{(1 / 2)}+C\)

= \(\frac{1}{2} \tan ^{-1}(2 t)+C\)=\(\frac{1}{2} \tan ^{-1}(2 \tan x)+C .\)

(2) Dividing the numerator and denominator by cos2x, we get

\(\int \frac{d x}{\left(3+2 \cos ^2 x\right)}=\int \frac{\sec ^2 x}{\left(3 \sec ^2 x+2\right)} d x=\int \frac{\sec ^2 x}{5+3 \tan ^2 x} d x\)

= \(\int \frac{d t}{\left(5+3 t^2\right)}\) [putting tan x = t]

= \(\frac{1}{3} \int \frac{d t}{\left[t^2+\left(\frac{5}{3}\right)\right]}=\frac{1}{3} \cdot \int \frac{d t}{\left[t^2+\left(\frac{\sqrt{5}}{\sqrt{3}}\right)^2\right]}\)

= \(\frac{1}{3} \cdot \frac{1}{\left(\frac{\sqrt{5}}{\sqrt{3}}\right)} \tan ^{-1} \frac{t}{\left(\frac{\sqrt{5}}{\sqrt{3}}\right)}+C=\frac{1}{\sqrt{15}} \tan ^{-1}\left(\frac{\sqrt{3} t}{\sqrt{5}}\right)+C\)

= \(\frac{1}{\sqrt{15}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{5}}\right)+C.\)

Class 12 Integration Special Integrals 

Example 25 Evaluate \(\int \frac{d x}{\left(4 \sin ^2 x+5 \cos ^2 x\right)}\).

Solution

\(\int \frac{d x}{\left(4 \sin ^2 x+5 \cos ^2 x\right)}\)

On dividing the numerator and denominator by cos2x, we get

\(\int \frac{d x}{\left(4 \sin ^2 x+5 \cos ^2 x\right)}=\int \frac{\sec ^2 x}{\left(4 \tan ^2 x+5\right)} d x\)

= \(\int \frac{d t}{4 t^2+5}\) [putting tan x = t]

= \(\frac{1}{4} \int \frac{d t}{\left(t^2+\frac{5}{4}\right)}=\frac{1}{4} \cdot \int \frac{d t}{\left[t^2+\left(\frac{\sqrt{5}}{2}\right)^2\right]}=\frac{1}{4} \cdot \frac{1}{\left(\frac{\sqrt{5}}{2}\right)} \tan ^{-1} \frac{t}{\left(\frac{\sqrt{5}}{2}\right)}+C\)

= \(\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 t}{\sqrt{5}}\right)+C=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+C\).

Example 26 Example \(\int \frac{d x}{\left(1+3 \sin ^2 x+8 \cos ^2 x\right)}\).

Solution

\(\int \frac{d x}{\left(1+3 \sin ^2 x+8 \cos ^2 x\right)}\)

On dividing the numerator and denominator by cos2x, we get

\(\int \frac{d x}{\left(1+3 \sin ^2 x+8 \cos ^2 x\right)}=\int \frac{\sec ^2 x d x}{\sec ^2 x+3 \tan ^2 x+8}\)

= \(\int \frac{\sec ^2 x}{9+4 \tan ^2 x} d x=\int \frac{d t}{9+4 t^2}\) [putting tan x = t]

= \(\frac{1}{4} \int \frac{d t}{\left(t^2+\frac{9}{4}\right)}=\frac{1}{4} \cdot \int \frac{d t}{\left[t^2+\left(\frac{3}{2}\right)^2\right]}\)

= \(\frac{1}{4} \cdot \frac{1}{(3 / 2)} \cdot \tan ^{-1}\left\{\frac{t}{(3 / 2)}\right\}+C\)

= \(\frac{1}{6} \tan ^{-1}\left(\frac{2 t}{3}\right)+C=\frac{1}{6} \tan ^{-1}\left(\frac{2 \tan x}{3}\right)+C\)

Techniques for Evaluating Special Integrals

Class 12 Integration Special Integrals 

Example 27 Evaluate \(\int \frac{\sin x}{\sin 3 x} d x\).

Solution We have

\(\int \frac{\sin x}{\sin 3 x} d x=\int \frac{\sin x}{\left(3 \sin x-4 \sin ^3 x\right)} d x\)

= \(\int \frac{1}{\left(3-4 \sin ^2 x\right)} d x\) [dividing num. and denom. by sin x]

= \(\int \frac{\sec ^2 x}{3 \sec ^2 x-4 \tan ^2 x} d x\) [dividing num. and denom. by cos2x]

= \(\int \frac{\sec ^2 x}{3\left(1+\tan ^2 x\right)-4 \tan ^2 x} d x=\int \frac{\sec ^2 x}{\left(3-\tan ^2 x\right)} d x\)

= \(\int \frac{d t}{\left(3-t^2\right)}\), where tan x = t and sec2x dx = dt

= \(\int \frac{d t}{\left[(\sqrt{3})^2-t^2\right]}=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C\)

= \(\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+C\)

Example 28 Evaluate \(\int \frac{\cos x}{\cos 3 x} d x\).

Solution We have

\(\int \frac{\cos x}{\cos 3 x} d x=\int \frac{\cos x}{\left(4 \cos ^3 x-3 \cos x\right)} d x\)

= \(\int \frac{d x}{4 \cos ^2 x-3}=\int \frac{d x}{4 \cos ^2 x-3\left(\sin ^2 x+\cos ^2 x\right)}\)

= \(\int \frac{d x}{\cos ^2 x-3 \sin ^2 x}=\int \frac{\sec ^2 x}{1-3 \tan ^2 x} d x\)

[on dividing the num. and denom. by cos2x]

= \(\int \frac{d t}{1-3 t^2}, where tan x = t and sex2x dx = dt\)

= \(\frac{1}{3} \cdot \int \frac{d t}{\left(\frac{1}{3}-t^2\right)}=\frac{1}{3} \cdot \int \frac{d t}{\left[\left(\frac{1}{\sqrt{3}}\right)^2-t^2\right]}=\frac{1}{3} \cdot \frac{1}{2 \cdot \frac{1}{\sqrt{3}}} \log \left|\frac{\frac{1}{\sqrt{3}}+t}{\frac{1}{\sqrt{3}}-t}\right|+C\)

= \(\frac{1}{2 \sqrt{3}} \log \left|\frac{1+\sqrt{3} t}{1-\sqrt{3} t}\right|+C=\frac{1}{2 \sqrt{3}} \log \left|\frac{1+\sqrt{3} \tan x}{1-\sqrt{3} \tan x}\right|+C .\)

Example 29 Evaluate \(\int \frac{d x}{(2+\cos x)}\)

Solution We have

\(\int \frac{d x}{(2+\cos x)}=\int \frac{d x}{1+(1+\cos x)}=\int \frac{d x}{1+2 \cos ^2(x / 2)}=\int \frac{\sec ^2(x / 2) d x}{\sec ^2(x / 2)+2}\)

[dividing the num. and denom. by cos2(x/2)]

= \(\int \frac{\sec ^2(x / 2)}{3+\tan ^2(x / 2)} d x=2 \int \frac{d t}{3+t^2}, \text { where } \tan (x / 2)=t\)

= \(2 \cdot \int \frac{d t}{(\sqrt{3})^2+t^2}=2 \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{t}{\sqrt{3}}+C\)

= \(\frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{\tan (x / 2)}{\sqrt{3}}\right]+C .\)

Special Integrals Class 12 Maths 

Some More Special Integrals

Example 30 Evaluate \(\int \frac{\left(x^2+1\right)}{\left(x^4+1\right)} d x\)

Solution We have

\(\int\left(\frac{x^2+1}{x^4+1}\right) d x=\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x^2}\right)} d x\) [dividing num. and denom. by x2]

= \(\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+2} d x\)

= \(\int \frac{d t}{\left[t^2+(\sqrt{2})^2\right]}\), wnere (x – \(\frac{1}{x}\)) = t and (1 + \(\frac{1}{x^2}\))dx = dt

= \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+C=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C\)

= \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)+C\)

Example 31 Evaluate \(\int \frac{\left(x^2+4\right)}{\left(x^4+16\right)} d x\)

Solution We have

\(\int\left(\frac{x^2+4}{x^4+16}\right) d x=\int \frac{\left(1+\frac{4}{x^2}\right)}{\left(x^2+\frac{16}{x^2}\right)} d x\) [dividing num. and denom. by x2]

= \(\int \frac{\left(1+\frac{4}{x^2}\right)}{\left(x-\frac{4}{x}\right)^2+8} d x\)

= \(\int \frac{d t}{\left(t^2+8\right)}\) [putting (x – \(\frac{4}{x}\)) = t and (1 + \(\frac{4}{x^2}\))dx = dt]

= \(\int \frac{d t}{t^2+(\sqrt{8})^2}\)=\(\frac{1}{\sqrt{8}} \tan ^{-1}\left(\frac{t}{\sqrt{8}}\right)+C\)

= \(\frac{1}{\sqrt{8}} \tan ^{-1} \frac{\left(x-\frac{4}{x}\right)}{\sqrt{8}}+C\)=\(\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2-4}{2 \sqrt{2} x}\right)+C\)

Special Integrals Class 12 Maths 

Example 32 Evaluate \(\int \frac{\left(x^2-1\right)}{\left(x^4+x^2+1\right)} d x\)

Solution We have

\(\int \frac{\left(x^2-1\right)}{\left(x^4+x^2+1\right)} d x=\int \frac{\left(1-\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x^2}+1\right)} d x\)

[dividing num. and denom. by x2]

= \(\int \frac{\left(1-\frac{1}{x^2}\right)}{\left[\left(x+\frac{1}{x}\right)^2-1\right]} d x\)

= \(\int \frac{d t}{\left(t^2-1\right)}\) [putting (x + \(\frac{1}{x}\)) = t and (1 – \(\frac{1}{x^2}\))dx = dt]

= \(\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+C=\frac{1}{2} \log \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|+C\)

= \(\frac{1}{2} \log \left|\frac{x^2-x+1}{x^2+x+1}\right|+C\)

Example 33 Evaluate \(\int \frac{d x}{\left(x^4+1\right)}\)

Solution We have

\(\int \frac{d x}{\left(x^4+1\right)}=\int \frac{\left(x^2+1\right)-\left(x^2-1\right)}{2\left(x^4+1\right)} d x\)

= \(\frac{1}{2} \int \frac{\left(x^2+1\right)}{\left(x^4+1\right)} d x-\frac{1}{2} \int \frac{\left(x^2-1\right)}{\left(x^4+1\right)} d x\)

= \(\frac{1}{2}\left[\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x^2}\right)} d x-\int \frac{\left(1-\frac{1}{x^2}\right)}{\left(x^2+\frac{1}{x^2}\right)} d x\right]\)

[dividing num. and denom. of each integral by x2]

= \(\frac{1}{2}\left[\int \frac{\left(1+\frac{1}{x^2}\right)}{\left[\left(x-\frac{1}{x}\right)^2+2\right]} d x-\int \frac{\left(1-\frac{1}{x^2}\right)}{\left[\left(x+\frac{1}{x}\right)^2-2\right]} d x\right]\)

= \(\frac{1}{2}\left[\int \frac{d t}{\left[t^2+(\sqrt{2})^2\right]}-\int \frac{d u}{\left[u^2-(\sqrt{2})^2\right]}\right]\)

[putting (x – \(\frac{1}{x}\)) = t in the 1st integral, and (x + \(\frac{1}{x}\)) = u in the 2nd]

= \(\frac{1}{2}\left\{\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|\right\}+C\)

= \(\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{4 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+C\)

= \(\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)-\frac{1}{4 \sqrt{2}} \log \left|\frac{x^2+1-\sqrt{2} x}{x^2+1+\sqrt{2 x}}\right|+C \text {. }\)

Real-Life Applications of Special Integrals

Example 34 Evaluate \(\int \frac{x^2}{\left(x^4+x^2+1\right)} d x\)

Solution We have

I = \(\frac{1}{2} \int \frac{2 x^2}{\left(x^4+x^2+1\right)} d x\)

= \(\frac{1}{2} \cdot \int \frac{\left(x^2-1\right)+\left(x^2+1\right)}{\left(x^4+x^2+1\right)} d x\)

= \(\frac{1}{2} \int \frac{\left(x^2-1\right)}{\left(x^4+x^2+1\right)} d x+\frac{1}{2} \int \frac{\left(x^2+1\right)}{\left(x^4+x^2+1\right)} d x\)

= \(\frac{1}{2} \int \frac{\left(1-\frac{1}{x^2}\right)}{\left(x^2+1+\frac{1}{x^2}\right)} d x+\frac{1}{2} \int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x^2+1+\frac{1}{x^2}\right)} d x\)

[on dividing num. and denom. of each by x2]

= \(\frac{1}{2} \int \frac{\left(1-\frac{1}{x^2}\right)}{\left\{\left(x+\frac{1}{x}\right)^2-1\right\}} d x+\frac{1}{2} \int \frac{\left(1+\frac{1}{x^2}\right)}{\left\{\left(x-\frac{1}{x}\right)^2+(\sqrt{3})^2\right\}} d x\)

{putting (x + \(\frac{1}{x}\)) = u and (1 – \(\frac{1}{x^2}\))dx = du in I1, and (x – \(\frac{1}{x}\)) = v and (1 + \(\frac{1}{x^2}\))dx = dv in I2}

= \(\frac{1}{2} \log \left|\frac{u-1}{u+1}\right|+\frac{1}{2} \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{v}{\sqrt{3}}+C\)

= \(\frac{1}{2} \log \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|+\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{3} x}\right)+C\)

= \(\frac{1}{2} \log \left|\frac{x^2-x+1}{x^2+x+1}\right|+\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{x^2+1}{\sqrt{3} x}\right)+C\)

Special Integrals Class 12 Maths 

Example 35 Evaluate \(\int \sqrt{\cot x} d x\)

Solution

\(\int \sqrt{\cot x} d x\)

Put cot x = t2 so that -cosec2x dx = 2t dt or dx = \(\frac{-2 t}{\left(1+t^4\right)} d t\).

∴ \(\int \sqrt{\cot x} d x=-\int \frac{2 t^2}{\left(t^4+1\right)} d t\)

= \(-\int \frac{\left[\left(t^2+1\right)+\left(t^2-1\right)\right]}{\left(t^4+1\right)} d t=-\int \frac{\left(t^2+1\right)}{\left(t^4+1\right)} d t-\int \frac{\left(t^2-1\right)}{\left(t^4+1\right)} d t\)

= \(-\int \frac{\left(1+\frac{1}{t^2}\right)}{\left(t^2+\frac{1}{t^2}\right)} d t-\int \frac{\left(1-\frac{1}{t^2}\right)}{\left(t^2+\frac{1}{t^2}\right)} d t\)

= \(-\int \frac{\left(1+\frac{1}{t^2}\right)}{\left[\left(t-\frac{1}{t}\right)^2+2\right]} d t-\int \frac{\left(1-\frac{1}{t^2}\right)}{\left[\left(t+\frac{1}{t}\right)^2-2\right]} d t\)

= \(-\int \frac{d u}{\left[u^2+(\sqrt{2})^2\right]}-\int \frac{d v}{\left[v^2-(\sqrt{2})^2\right]}\)

[putting (t – \(\frac{1}{t}\)) = u in the 1st integral, and (t + \(\frac{1}{x}\)) = v in the 2nd]

= \(-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right|+C\)

= \(-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+C\)

= \(-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t^2-1}{\sqrt{2} t}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{t^2-\sqrt{2} t+1}{t^2+\sqrt{2} t+1}\right|+C\)

= \(-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\cot x-1}{\sqrt{2 \cot x}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{\cot x-\sqrt{2 \cot x}+1}{\cot x+\sqrt{2 \cot x}+1}\right|+C\)

Example 36 Evaluate \(\int(\sqrt{\tan x}+\sqrt{\cot x}) d x\)

Solution We have

\(\int(\sqrt{\tan x}+\sqrt{\cot x}) d x\)

= \(\int\left(\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}\right) d x=\int \frac{(\tan x+1)}{\sqrt{\tan x}} d x\)

= \(\int \frac{\left(t^2+1\right)}{t} \cdot \frac{2 t}{\left(1+t^4\right)} d t\), where tan x = t2 ⇒ x = tan-1t2

⇒ \(d x=\frac{2 t}{\left(1+t^4\right)} d t\)

= \(2 \int \frac{\left(t^2+1\right)}{\left(t^4+1\right)} d t=2 \int \frac{\left(1+\frac{1}{t^2}\right)}{\left(t^2+\frac{1}{t^2}\right)} d t\) [on dividing num. and denom. by t2]

= \(2 \int \frac{\left(1+\frac{1}{t^2}\right)}{\left(t-\frac{1}{t}\right)^2+2} d t\)

= \(2 \int \frac{d u}{\left(u^2+2\right)}, where (t – [latex]\frac{1}{t}\)) = u and (1 + \(\frac{1}{t^2}\))dt = du

= \(2 \cdot \frac{1}{\sqrt{2}} \tan ^{-1} \frac{u}{\sqrt{2}}+C=\sqrt{2} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+C\) [∵ u = (t – \(\frac{1}{t}\))]

= \(\sqrt{2} \tan ^{-1} \frac{\left(t^2-1\right)}{(\sqrt{2} t)}+C=\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+C\) [∵ t2 = tan x].

Example 37 Evaluate \(\int \sqrt{\tan \theta} d \theta\).

Solution

\(\int \sqrt{\tan \theta} d \theta\)

Putting tan θ = t2, we get θ = tan-1t2 ⇒ \(d \theta=\frac{2 t}{\left(1+t^4\right)} d t .\)

∴ I = \(\int t \cdot \frac{2 t}{\left(1+t^4\right)} d t=\int \frac{2 t^2}{\left(t^4+1\right)} d t\)

= \(\int \frac{\left(t^2+1\right)+\left(t^2-1\right)}{\left(t^4+1\right)} d t=\int \frac{\left(t^2+1\right)}{\left(t^4+1\right)} d t+\int \frac{\left(t^2-1\right)}{\left(t^4+1\right)} d t\)

= \(\int \frac{\left(1+\frac{1}{t^2}\right)}{\left(t^2+\frac{1}{t^2}\right)} d t+\int \frac{\left(1-\frac{1}{t^2}\right)}{\left(t^2+\frac{1}{t^2}\right)} d t\)

= \(\int \frac{\left(1+\frac{1}{t^2}\right)}{\left(t-\frac{1}{t}\right)^2+2} d t+\int \frac{\left(1-\frac{1}{t^2}\right)}{\left\{\left(t+\frac{1}{t}\right)^2-2\right\}} d t\)

= \(\int \frac{d u}{\left(u^2+2\right)}+\int \frac{d v}{\left(v^2-2\right)}\),

where (t – \(\frac{1}{t}\)) = u and (t + \(\frac{1}{t}\)) = v in I1 and I2

= \(\frac{1}{\sqrt{2}} \tan ^{-1} \frac{u}{\sqrt{2}}+\frac{1}{2 \sqrt{2}} \cdot \log \left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right|+C\)

= \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+\frac{1}{2 \sqrt{2}} \cdot\left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+C\)

= \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t^2-1}{\sqrt{2} t}\right)+\frac{1}{2 \sqrt{2}} \cdot\left|\frac{t^2-\sqrt{2} t+1}{t^2+\sqrt{2} t+1}\right|+C\), where t = \(\sqrt{\tan \theta}\).

Special Integrals Class 12 Maths 

Three More Special Integrals

Theorem

(1) \(\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C .\)

(2) \(\int \frac{d x}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C\).

(3) \(\int \frac{d x}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C\).

Proof

(1) Put x = a sin θ so that dx = a cos θ dθ.

∴ \(\int \frac{d x}{\sqrt{a^2-x^2}}=\int \frac{a \cos \theta}{a \cos \theta} d \theta=\int d \theta=\theta+C=\sin ^{-1} \frac{x}{a}+C\).

Hence, \(\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C\)

(2) Put x = a sec θ so that dx = a sec θ tan θ dθ.

∴ \(\int \frac{d x}{\sqrt{x^2-a^2}}=\int \frac{a \sec \theta \tan \theta}{a \tan \theta} d \theta\)

= \(\int \sec \theta d \theta=\log |\sec \theta+\tan \theta|+c\)

= \(\log \left|\sec \theta+\sqrt{\sec ^2 \theta-1}\right|+c\)

= \(\log \left|\frac{x}{a}+\sqrt{\left(\frac{x^2}{a^2}-1\right)}\right|+c=\log \left|\frac{x+\sqrt{x^2-a^2}}{a}\right|+c\)

= \(\log \left|x+\sqrt{x^2-a^2}\right|-\log a+c\)

= \(\log \left|x+\sqrt{x^2-a^2}\right|+C\) [taking -log a + c = C].

Hence, \(\int \frac{d x}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C\).

(3) Put x = a tan θ so that dx = a sec2θ dθ.

∴ \(\int \frac{d x}{\sqrt{x^2+a^2}}=\int \frac{a \sec ^2 \theta}{a \sec \theta} d \theta\)

= \(\int \sec \theta d \theta=\log |\sec \theta+\tan \theta|+c\)

= \(\log \left|\sqrt{1+\tan ^2 \theta}+\tan \theta\right|+c\)

= \(\log \left|\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}\right|+c\)

= \(\log \left|x+\sqrt{x^2+a^2}\right|-\log a+c\)

= \(\log \left|x+\sqrt{x^2+a^2}\right|+C\) [taking -log a + c = C].

Hence, \(\int \frac{d x}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C .\)

Class 12 Maths Special Integrals Solutions 

Solved Examples

Example 1 Evaluate:

(1) \(\int \frac{d x}{\sqrt{9-25 x^2}}\)

(2) \(\int \frac{d x}{\sqrt{4 x^2-9}}\)

(3) \(\int \frac{d x}{\sqrt{16 x^2+25}}\)

Solution We have

(1) \(\int \frac{d x}{\sqrt{9-25 x^2}}=\frac{1}{5} \cdot \int \frac{d x}{\sqrt{\frac{9}{25}-x^2}}=\frac{1}{5} \cdot \int \frac{d x}{\sqrt{\left(\frac{3}{5}\right)^2-x^2}}\)

= \(\frac{1}{5} \sin ^{-1}\left(\frac{x}{3 / 5}\right)+C=\frac{1}{5} \sin ^{-1}\left(\frac{5 x}{3}\right)+C .\)

(2) \(\int \frac{d x}{\sqrt{4 x^2-9}}=\frac{1}{2} \int \frac{d x}{\sqrt{x^2-\frac{9}{4}}}=\frac{1}{2} \int \frac{d x}{\sqrt{x^2-\left(\frac{3}{2}\right)^2}}\)

= \(\frac{1}{2} \log \left|x+\sqrt{x^2-\frac{9}{4}}\right|+C\)

= \(\frac{1}{2} \log \left|2 x+\sqrt{4 x^2-9}\right|+C\).

(3) \(\int \frac{d x}{\sqrt{16 x^2+25}}=\frac{1}{4} \cdot \int \frac{d x}{\sqrt{x^2+\frac{25}{16}}}=\frac{1}{4} \cdot \int \frac{d x}{\sqrt{x^2+\left(\frac{5}{4}\right)^2}}\)

= \(\frac{1}{4} \cdot \log \left|x+\sqrt{x^2+\frac{25}{16}}\right|+C\)

= \(\frac{1}{4} \log \left|4 x+\sqrt{16 x^2+25}\right|+C .\)

Example 2 Evaluate \(\int \frac{d x}{\sqrt{15-8 x^2}}\).

Solution We have

\(\int \frac{d x}{\sqrt{15-8 x^2}}=\frac{1}{\sqrt{8}} \cdot \int \frac{d x}{\sqrt{\frac{15}{8}-x^2}}\)

= \(\frac{1}{2 \sqrt{2}} \sin ^{-1}\left\{\frac{x}{\sqrt{\frac{15}{8}}}\right\}+C=\frac{1}{2 \sqrt{2}} \sin ^{-1}\left(\sqrt{\frac{8}{15}} x\right)+C\).

Example 3 Evaluate \(\int \frac{\cos x}{\sqrt{4-\sin ^2 x}} d x\)

Solution

\(\int \frac{\cos x}{\sqrt{4-\sin ^2 x}} d x\)

Putting sin x = t and cos x dx = dt, we get

\(\int \frac{\cos x}{\sqrt{4-\sin ^2 x}} d x=\int \frac{d t}{\sqrt{4-t^2}}=\int \frac{d t}{\sqrt{2^2-t^2}}\)

= \(\frac{1}{2 \sqrt{2}} \sin ^{-1}\left\{\frac{x}{\sqrt{\frac{15}{8}}}\right\}+C=\frac{1}{2 \sqrt{2}} \sin ^{-1}\left(\sqrt{\frac{8}{15} x}\right)+C.\)

Class 12 Maths Special Integrals Solutions 

Example 4 Evaluate \(\int \frac{d x}{\sqrt{1-e^{2 x}}} .\)

Solution

\(\int \frac{d x}{\sqrt{1-e^{2 x}}} .\)

Multiplying the numerator and denominator by e-x, we get

\(\int \frac{d x}{\sqrt{1-e^{2 x}}}=\int \frac{e^{-x} d x}{\sqrt{e^{-2 x}\left(1-e^{2 x}\right)}}=\int \frac{e^{-x}}{\sqrt{e^{-2 x}-1}} d x\)

= \(-\int \frac{d t}{\sqrt{t^2-1}}\) [putting e-x = t]

= \(-\log \left|t+\sqrt{t^2-1}\right|+C\)

= \(-\log \left|e^{-x}+\sqrt{e^{-2 x}-1}\right|+C .\)

Example 5 Evaluate \(\int \frac{2^x}{\sqrt{1-4^x}} d x\).

Solution

\(\int \frac{2^x}{\sqrt{1-4^x}} d x\)

Putting 2x = t and (2x log 2)dx = dt, we get

\(\int \frac{2^x}{\sqrt{1-4^x}} d x=\frac{1}{(\log 2)} \cdot \int \frac{d t}{\sqrt{1-t^2}}\)

= \(\frac{1}{(\log 2)} \cdot \sin ^{-1} t+C=\frac{1}{(\log 2)} \cdot \sin ^{-1}\left(2^x\right)+C .\)

Example 6  Evaluate \(\int \frac{x^2}{\sqrt{x^6-1}} d x\)

Solution

\(\int \frac{x^2}{\sqrt{x^6-1}} d x\)

Putting x3 = t and x2dx = \(\frac{1}{3}\) dt, we get

\(\int \frac{x^2}{\sqrt{x^6-1}} d x=\frac{1}{3} \int \frac{d t}{\sqrt{t^2-1}}=\frac{1}{3} \log \left|t+\sqrt{t^2-1}\right|+C\)

= \(\frac{1}{3} \log \left|x^3+\sqrt{x^6-1}\right|+C\).

Example 7 Evaluate:

(1) \(\int \frac{\sin x}{\sqrt{4 \cos ^2 x-1}} d x\)

(2) \(\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x-4}} d x\)

Solution

(1) Putting cos x = t and -sin x dx = dt, we get

\(\int \frac{\sin x}{\sqrt{4 \cos ^2 x-1}} d x=\int \frac{-d t}{\sqrt{4 t^2-1}}=-\frac{1}{2} \int \frac{d t}{\sqrt{t^2-(1 / 2)^2}}\)

= \(-\frac{1}{2} \cdot \log \left|t+\sqrt{t^2-\frac{1}{4}}\right|+C\)

= \(-\frac{1}{2} \log \left|2 t+\sqrt{4 t^2-1}\right|+C\)

= \(-\frac{1}{2} \log \left|2 \cos x+\sqrt{4 \cos ^2 x-1}\right|+C \text {. }\)

(2) Putting tan x = t and sec2xdx = dt, we get

\(\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x-4}} d x=\int \frac{d t}{\sqrt{t^2-4}}=\log \left|t+\sqrt{t^2-4}\right|+C\)

= \(\log \left|\tan x+\sqrt{\tan ^2 x-4}\right|+C\).

Class 12 Maths Special Integrals Solutions 

Example 8 Evaluate \(\int \frac{x^2}{\sqrt{x^6+a^6}} d x\)

Solution

\(\int \frac{x^2}{\sqrt{x^6+a^6}} d x\)

Putting x3 = t and x2dx = \(\frac{1}{3}\)dt, we get

\(\int \frac{x^2}{\sqrt{x^6+a^6}} d x=\frac{1}{3} \int \frac{d t}{\sqrt{t^2+\left(a^3\right)^2}}=\frac{1}{3} \log \left|t+\sqrt{t^2+a^6}\right|+C\)

= \(\frac{1}{3} \log \left|x^3+\sqrt{x^6+a^6}\right|+C .\)

Example 9 Evaluate \(\int \frac{\sec ^2 x}{\sqrt{16+\tan ^2 x}} d x.\)

Solution

\(\int \frac{\sec ^2 x}{\sqrt{16+\tan ^2 x}} d x.\)

Putting tan x = t and sec2xdx = dt, we get

\(\int \frac{\sec ^2 x}{\sqrt{16+\tan ^2 x}} d x\)=\(\int \frac{d t}{\sqrt{16+t^2}}=\log \left|t+\sqrt{t^2+16}\right|+C\)

= \(\log \left|\tan x+\sqrt{\tan ^2 x+16}\right|+C .\)

Example 10 Evaluate \(\int \sqrt{\frac{1-x}{1+x}} d x\).

Solution We have

\(\int \sqrt{\frac{1-x}{1+x}} d x=\int\left\{\frac{\sqrt{1-x}}{\sqrt{1+x}} \times \frac{\sqrt{1-x}}{\sqrt{1-x}}\right\} d x\)

= \(\int \frac{(1-x)}{\sqrt{1-x^2}} d x=\int \frac{d x}{\sqrt{1-x^2}}-\int \frac{x}{\sqrt{1-x^2}} d x\)

= \(\sin ^{-1} x+\frac{1}{2} \cdot \int \frac{(-2 x)}{\sqrt{1-x^2}} d x\)

= \(\sin ^{-1} x+\frac{1}{2} \int \frac{d t}{\sqrt{t}}\), where (1 – x2) = t and (-2x)dx = dt

= \(\sin ^{-1} x+\frac{1}{2} \int t^{-1 / 2} d t=\sin ^{-1} x+\frac{1}{2} \cdot \frac{t^{1 / 2}}{(1 / 2)}+C\)

= \(\sin ^{-1} x+\sqrt{1-x^2}+C .\)

Integrals of the form \(\int \frac{d x}{\sqrt{\left(a x^2+b x+c\right)}} .\)

Method Put (ax2 + bx + c) in the form a{(x + α)2 ± β2} and then integrate.

Example 11 Evaluate \(\int \frac{d x}{\sqrt{x^2-3 x+2}}\).

Solution We have

\(\int \frac{d x}{\sqrt{x^2-3 x+2}}=\int \frac{d x}{\sqrt{\left(x^2-3 x+\frac{9}{4}\right)-\frac{1}{4}}}=\int \frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}}\)

= \(\int \frac{d z}{\sqrt{z^2-\left(\frac{1}{2}\right)^2}}\), where (x – \(\frac{3}{2}\)) = z

= \(\log \left|z+\sqrt{z^2-\frac{1}{4}}\right|+C\)

[∵ \(\int \frac{d z}{\sqrt{z^2-a^2}}=\log \left|z+\sqrt{z^2-a^2}\right|+C\)]

= \(\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+C\).

Class 12 Maths Special Integrals Solutions 

Example 12 Evaluate \(\int \frac{d x}{\sqrt{5 x^2-2 x}}\).

Solution We have

\(\int \frac{d x}{\sqrt{5 x^2-2 x}}=\frac{1}{\sqrt{5}} \cdot \int \frac{d x}{\sqrt{x^2-\frac{2}{5} x}}=\frac{1}{\sqrt{5}} \cdot \int \frac{d x}{\sqrt{x^2-\frac{2}{5} x+\left(\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}}\)

= \(\frac{1}{\sqrt{5}} \cdot \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}}=\frac{1}{\sqrt{5}} \cdot \int \frac{d t}{\sqrt{t^2-\left(\frac{1}{5}\right)^2}}\)

where (x – \(\frac{1}{5}\)) = t

= \(\frac{1}{\sqrt{5}} \cdot \log \left|t+\sqrt{t^2-\left(\frac{1}{5}\right)^2}\right|+C\)

= \(\frac{1}{\sqrt{5}} \log \left|\left(x-\frac{1}{5}\right)+\sqrt{x^2-\frac{2 x}{5}}\right|+C\)

Example 13 Evaluate \(\int \frac{\cos x}{\sqrt{\sin ^2 x-2 \sin x-3}} d x .\)

Solution

\(\int \frac{\cos x}{\sqrt{\sin ^2 x-2 \sin x-3}} d x .\)

Putting sin x = t and cos x dx = dt, we get

\(\int \frac{\cos x}{\sqrt{\sin ^2 x-2 \sin x-3}} d x=\int \frac{d t}{\sqrt{t^2-2 t-3}}=\int \frac{d t}{\sqrt{(t-1)^2-2^2}}\)

= \(\log \left|(t-1)+\sqrt{(t-1)^2-2^2}\right|+C\)

= \(\log \left|(\sin x-1)+\sqrt{\sin ^2 x-2 \sin x-3}\right|+C \text {. }\)

Example 14 Evaluate \(\int \frac{e^x}{\sqrt{5-4 e^x-e^{2 x}}} d x\).

Solution

\(\int \frac{e^x}{\sqrt{5-4 e^x-e^{2 x}}} d x\)

Putting ex = t and exdx = dt, we get

\(\int \frac{e^x}{\sqrt{5-4 e^x-e^{2 x}}} d x=\int \frac{d t}{\sqrt{5-4 t-t^2}}=\int \frac{d t}{\sqrt{5-\left(t^2+4 t+4\right)+4}}\)

= \(\int \frac{d t}{\sqrt{9-(t+2)^2}}=\int \frac{d t}{\sqrt{3^2-(t+2)^2}}\)

= \(\int \frac{d z}{\sqrt{3^2-z^2}}\), where (t+2) = z

= \(\sin ^{-1} \frac{z}{3}+C=\sin ^{-1} \frac{(t+2)}{3}+C\)

= \(\sin ^{-1} \frac{\left(e^x+2\right)}{3}+C\)

Example 15 Evaluate \(\int \frac{d x}{\sqrt{2-4 x+x^2}}\)

Solution We have

\(\int \frac{d x}{\sqrt{2-4 x+x^2}}=\int \frac{d x}{\sqrt{x^2-4 x+4-2}}=\int \frac{d x}{\sqrt{(x-2)^2-(\sqrt{2})^2}}\)

= \(\log \left|(x-2)+\sqrt{(x-2)^2-2}\right|+C\)

= \(\log \left|x-2+\sqrt{x^2-4 x+2}\right|+C\).

Example 16 Evaluate \(\int \frac{d x}{\sqrt{3 x^2+6 x+12}}\)

Solution We have

\(\int \frac{d x}{\sqrt{3 x^2+6 x+12}}=\frac{1}{\sqrt{3}} \cdot \int \frac{d x}{\sqrt{x^2+2 x+4}}\)

= \(\frac{1}{\sqrt{3}} \cdot \int \frac{d x}{\sqrt{(x+1)^2+(\sqrt{3})^2}}\)

= \(\frac{1}{\sqrt{3}} \cdot \int \frac{d t}{\sqrt{t^2+(\sqrt{3})^2}}\), where (x+1) = t

= \(\frac{1}{\sqrt{3}} \log \left|t+\sqrt{t^2+3}\right|+C\)

= \(\frac{1}{\sqrt{3}} \log \left|(x+1)+\sqrt{x^2+2 x+4}\right|+C\).

WBCHSE Maths Chapter Special Integrals 

Example 17 Evaluate \(\int \frac{d x}{\sqrt{8+3 x-x^2}}\)

Solution We have

\(\int \frac{d x}{\sqrt{8+3 x-x^2}}=\int \frac{d x}{\sqrt{8-\left(x^2-3 x\right)}}=\int \frac{d x}{\sqrt{8-\left(x^2-3 x+\frac{9}{4}\right)+\frac{9}{4}}}\)

= \(\int \frac{d x}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}=\sin ^{-1}\left\{\frac{\left(x-\frac{3}{2}\right)}{\left(\frac{\sqrt{41}}{2}\right)}\right\}+C\)

= \(\sin ^{-1}\left(\frac{2 x-3}{\sqrt{41}}\right)+C\).

Example 18 Evaluate \(\int \frac{d x}{\sqrt{2 x-x^2}} .\)

Solution We have

\(\int \frac{d x}{\sqrt{2 x-x^2}}=\int \frac{d x}{\sqrt{1-\left(x^2-2 x+1\right)}}\)

= \(\int \frac{d x}{\sqrt{1-(x-1)^2}}=\sin ^{-1}(x-1)+C \text {. }\)

Example 19 Evaluate \(\int \frac{d x}{\sqrt{x(1-2 x)}}\)

Solution We have

\(\int \frac{d x}{\sqrt{x(1-2 x)}}=\int \frac{d x}{\sqrt{x-2 x^2}}\)

= \(\frac{1}{\sqrt{2}} \cdot \int \frac{d x}{\sqrt{\frac{x}{2}-x^2}}=\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{-\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{1}{16}}}\)

= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{1}{16}-\left\{x^2-\frac{x}{2}+\frac{1}{16}\right\}}}=\frac{1}{\sqrt{2}} \cdot \int \frac{d x}{\sqrt{\left(\frac{1}{4}\right)^2-\left(x-\frac{1}{4}\right)^2}}\)

= \(\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{\left(\frac{1}{4}\right)^2-t^2}}\), where (x – \(\frac{1}{4}\)) = t

= \(\frac{1}{\sqrt{2}} \sin ^{-1} \frac{t}{(1 / 4)}+C=\frac{1}{\sqrt{2}} \sin ^{-1}(4 t)+C=\frac{1}{\sqrt{2}} \sin ^{-1} 4\left(x-\frac{1}{4}\right)+C\)

= \(\frac{1}{\sqrt{2}} \sin ^{-1}(4 x-1)+C .\)

Integrals of the from \(\int \frac{(p x+q)}{\sqrt{\left(a x^2+b x+c\right)}} d x\)

Method Let (px + q) = \(A \cdot \frac{d}{d x}\left(a x^2+b x+c\right)+B\).

Now, the value of the integral can be obtained easily.

Example 20 Evaluate \(\int \frac{2 x+1}{\sqrt{x^2+2 x-1}} d x .\)

Solution

\(\int \frac{2 x+1}{\sqrt{x^2+2 x-1}} d x .\)

Let 2x + 1 = \(A \cdot \frac{d}{d x}\left(x^2+2 x-1\right)+B\)

or 2x + 1 = A(2x+2) + B.

Comparing the coefficients of like powers of x, we get

(2A = 1 and 2A + B = 1) ⇒ (A = 1 and B = -1).

∴ \(\int \frac{2 x+1}{\sqrt{x^2+2 x-1}} d x=\int \frac{1 \cdot(2 x+2)-1}{\sqrt{x^2+2 x-1}} d x\)

= \(\int \frac{2 x+2}{\sqrt{x^2+2 x-1}} d x-\int \frac{d x}{\sqrt{x^2+2 x-1}}\)

= \(\int \frac{1}{\sqrt{t}} d t-\int \frac{d x}{\sqrt{x^2+2 x+1-2}}\), where t = x2 + 2x – 1

= \(2 \sqrt{t}-\int \frac{d x}{\sqrt{(x+1)^2-(\sqrt{2})^2}}\)

= \(2 \sqrt{x^2+2 x-1}-\log \left|(x+1)+\sqrt{x^2+2 x-1}\right|+C\)

WBCHSE Maths Chapter Special Integrals 

Example 21 Evaluate \(\int \frac{(x+1)}{\sqrt{4+5 x-x^2}} d x .\)

Solution

\(\int \frac{(x+1)}{\sqrt{4+5 x-x^2}} d x .\)

Let (x+1) = \(A \cdot \frac{d}{d x}\left(4+5 x-x^2\right)+B\). Then,

(x+1) = A(5-2x) + B.

Comparing the coefficients of like powers of x, we get

(-2A = 1 and 5A + B = 1) ⇒ (A = \(-\frac{1}{2}\) and B = \(\frac{7}{2}\)).

∴ (x+1) = \(-\frac{1}{2}\)(5-2x) + \(\frac{7}{2}\).

∴ \(\int \frac{(x+1)}{\sqrt{4+5 x-x^2}} d x=\int \frac{-\frac{1}{2}(5-2 x)+\frac{7}{2}}{\sqrt{4+5 x-x^2}} d x\)

= \(-\frac{1}{2} \int \frac{(5-2 x)}{\sqrt{4+5 x-x^2}} d x+\frac{7}{2} \int \frac{1}{\sqrt{4+5 x-x^2}} d x\)

= \(-\frac{1}{2} \int \frac{1}{\sqrt{t}} d t+\frac{7}{2} \cdot \int \frac{1}{\sqrt{4-\left(x^2-5 x\right)}} d x\), where t = 4 + 5x – x2

= \(-\frac{1}{2} \cdot 2 \sqrt{t}+\frac{7}{2} \cdot \int \frac{d x}{\sqrt{4-\left(x^2-5 x+\frac{25}{4}\right)+\frac{25}{4}}}\)

= \(-\sqrt{t}+\frac{7}{2} \cdot \int \frac{d x}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac{5}{2}\right)^2}}\)

= \(-\sqrt{4+5 x-x^2}+\frac{7}{2} \cdot \sin ^{-1} \frac{\left(x-\frac{5}{2}\right)}{\left(\frac{\sqrt{41}}{2}\right)}+C\)

= \(-\sqrt{4+5 x-x^2}+\frac{7}{2} \sin ^{-1}\left(\frac{2 x-5}{\sqrt{41}}\right)+C .\)

Practice Problems on Special Integrals for Class 12

Example 22 Evaluate \(\int \frac{(x+3)}{\sqrt{5-4 x-x^2}} d x\).

Solution

\(\int \frac{(x+3)}{\sqrt{5-4 x-x^2}} d x\)

Let (x+3) = \(A \cdot \frac{d}{d x}\left(5-4 x-x^2\right)+B\). Then,

(x+3) = A(-4-2x) + B.

Comparing the coefficients of like powers of x, we get

(-2A = 1 and -4A + B = 3) ⇒ (A = \(-\frac{1}{2}\), B = 1).

∴ \(\int \frac{(x+3)}{\sqrt{5-4 x-x^2}} d x=-\frac{1}{2} \cdot \int \frac{(-4-2 x)}{\sqrt{5-4 x-x^2}} d x+\int \frac{d x}{\sqrt{5-4 x-x^2}}\)

= \(-\frac{1}{2} \int \frac{d t}{\sqrt{t}}+\int \frac{d x}{\sqrt{5-\left(x^2+4 x+4\right)+4}}\), where (5-4x-x2) = t

= \(-\frac{1}{2} \cdot \frac{t^{1 / 2}}{(1 / 2)}+\int \frac{d x}{\sqrt{3^2-(x+2)^2}}\)

= \(-\sqrt{t}+\sin ^{-1} \frac{(x+2)}{3}+C\)

= \(-\sqrt{5-4 x-x^2}+\sin ^{-1} \frac{(x+2)}{3}+C\)

WBCHSE Maths Chapter Special Integrals 

Three More Special Integrals

Theorem

(1) \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)

(2) \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|+C\)

(3) \(\int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|+C .\)

Proof (1) Integrating by parts, taking 1 as the second function, we get

I = \(\int \sqrt{a^2-x^2} d x=\int\left(\sqrt{a^2-x^2} \cdot 1\right) d x\)

= \(\left(\sqrt{a^2-x^2}\right) \cdot x-\int \frac{1}{2}\left(a^2-x^2\right)^{-1 / 2}(-2 x) \cdot x d x\)

= \(x\left(\sqrt{a^2-x^2}\right)+\int \frac{x^2}{\sqrt{a^2-x^2}} d x=x\left(\sqrt{a^2-x^2}\right)+\int \frac{a^2-\left(a^2-x^2\right)}{\sqrt{a^2-x^2}} d x\)

= \(x\left(\sqrt{a^2-x^2}\right)+a^2 \int \frac{d x}{\sqrt{a^2-x^2}}-\int \sqrt{a^2-x^2} d x\)

= \(x\left(\sqrt{a^2-x^2}\right)+a^2 \sin ^{-1} \frac{x}{a}-I+c .\)

∴ \(2 I=x\left(\sqrt{a^2-x^2}\right)+a^2 \sin ^{-1} \frac{x}{a}+c\)

⇒ \(I=\frac{x}{2}\left(\sqrt{a^2-x^2}\right)+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\) [taking \frac{c}{2} = C].

Hence, \(\int \sqrt{a^2-x^2} d x=\frac{x}{2}\left(\sqrt{a^2-x^2}\right)+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+\mathrm{C} .\)

(2) Integrating by parts, taking 1 as the second function, we get

I = \(\int \sqrt{x^2-a^2} d x=\int\left(\sqrt{x^2-a^2} \cdot 1\right) d x\)

= \(\left(\sqrt{x^2-a^2}\right) \cdot x-\int \frac{1}{2}\left(x^2-a^2\right)^{-1 / 2}(2 x) \cdot x d x\)

= \(x\left(\sqrt{x^2-a^2}\right)-\int \frac{x^2}{\sqrt{x^2-a^2}} d x=x\left(\sqrt{x^2-a^2}\right)-\int \frac{\left(x^2-a^2\right)+a^2}{\sqrt{x^2-a^2}} d x\)

= \(x\left(\sqrt{x^2-a^2}\right)-\int \sqrt{x^2-a^2} d x-\int \frac{a^2}{\sqrt{x^2-a^2}} d x\)

= \(x\left(\sqrt{x^2-a^2}\right)-I-a^2 \log \left|x+\sqrt{x^2-a^2}\right|+c .\)

∴ \(2 I=x\left(\sqrt{x^2-a^2}\right)-a^2 \log \left|x+\sqrt{x^2-a^2}\right|+c\)

⇒ \(I=\frac{x}{2}\left(\sqrt{x^2-a^2}\right)-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|+C\) [taking \frac{c}{2} = C].

Hence, \(\int \sqrt{x^2-a^2} d x=\frac{x}{2}\left(\sqrt{x^2-a^2}\right)-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|+C .\)

(3) Integrating by parts, taking 1 as the second function, we get

I = \(\int \sqrt{x^2+a^2} d x=\int\left(\sqrt{x^2+a^2} \cdot 1\right) d x\)

= \(\left(\sqrt{x^2+a^2}\right) x-\int \frac{1}{2}\left(x^2+a^2\right)^{-1 / 2}(2 x) \cdot x d x\)

= \(x\left(\sqrt{x^2+a^2}\right)-\int \frac{x^2}{\sqrt{x^2+a^2}} d x=x\left(\sqrt{x^2+a^2}\right)-\int \frac{\left(x^2+a^2\right)-a^2}{\sqrt{x^2+a^2}} d x\)

= \(x\left(\sqrt{x^2+a^2}\right)-\int \sqrt{x^2+a^2} d x+a^2 \int \frac{d x}{\sqrt{x^2+a^2}}\)

= \(x \cdot\left(\sqrt{x^2+a^2}\right)-I+a^2 \log \left|x+\sqrt{x^2+a^2}\right|+c .\)

∴ \(2 I=x\left(\sqrt{x^2+a^2}\right)+a^2 \log \left|x+\sqrt{x^2+a^2}\right|+c\)

or I = \(\frac{x}{2}\left(\sqrt{x^2+a^2}\right)+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|+C\) [taking \frac{c}{2} = C].

Hence, \(\int \sqrt{x^2+a^2} d x=\frac{x}{2}\left(\sqrt{x^2+a^2}\right)+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|+C .\)

WBCHSE Maths Chapter Special Integrals 

Solved Examples

Example 1 Evaluate:

(1) \(\int \sqrt{9-x^2} d x\)

(2) \(\int \sqrt{1-4 x^2} d x\)

(3) \(\int \sqrt{16-9 x^2} d x\)

Solution

We know that \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)

∴ (1) \(\int \sqrt{9-x^2} d x=\int \sqrt{3^2-x^2} d x\)

= \(\frac{x}{2} \sqrt{9-x^2}+\frac{9}{2} \sin ^{-1} \frac{x}{3}+C .\)

(2) \(\sqrt{1-4 x^2} d x=2 \int \sqrt{\left(\frac{1}{4}-x^2\right)} d x=2 \int\left\{\sqrt{\left(\frac{1}{2}\right)^2-x^2}\right\} d x\)

= \(2\left[\frac{x}{2} \sqrt{\frac{1}{4}-x^2}+\frac{1}{8} \sin ^{-1}\left(\frac{x}{(1 / 2)}\right)\right]+C\)

= \(\frac{x}{2} \sqrt{1-4 x^2}+\frac{1}{4} \sin ^{-1}(2 x)+C .\)

(3) \(\int \sqrt{16-9 x^2} d x=3 \int\left\{\sqrt{\left(\frac{16}{9}-x^2\right)}\right\} d x=3 \int\left\{\sqrt{\left(\frac{4}{3}\right)^2-x^2}\right\} d x\)

= \(3\left[\frac{x}{2} \sqrt{\frac{16}{9}-x^2}+\frac{8}{9} \sin ^{-1} \frac{x}{(4 / 3)}\right]+C\)

= \(\frac{x}{2} \sqrt{16-9 x^2}+\frac{8}{3} \sin ^{-1}\left(\frac{3 x}{4}\right)+C .\)

Example 2 Evaluate:

(1) \(\int \sqrt{x^2-16} d x\)

(2) \(\int \sqrt{4 x^2-5} d x\)

(3) \(\int \sqrt{17 x^2-11} d x\)

Solution

We know that \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|+C .\)

∴ (1) \(\int \sqrt{x^2-16} d x=\int \sqrt{x^2-4^2} d x\)

= \(\frac{x}{2} \cdot \sqrt{x^2-4^2}-\frac{16}{2} \log \left|x+\sqrt{x^2-16}\right|+C\)

= \(\frac{x}{2} \cdot \sqrt{x^2-16}-8 \log \left|x+\sqrt{x^2-16}\right|+C\)

(2) \(\int \sqrt{4 x^2-5} d x=2 \int \sqrt{x^2-\frac{5}{4}} d x=2 \cdot \int \sqrt{x^2-\left(\frac{\sqrt{5}}{2}\right)^2} d x\)

= \(\left[\frac{x}{2} \sqrt{x^2-\frac{5}{4}}-\frac{5}{8} \log \left|x+\sqrt{x^2-\frac{5}{4}}\right|\right]+C\)

= \(x \sqrt{x^2-\frac{5}{4}}-\frac{5}{4} \log \left|x+\sqrt{x^2-\frac{5}{4}}\right|+C\)

(3) \(\int \sqrt{17 x^2-11} d x=\sqrt{17} \cdot \int \sqrt{x^2-\frac{11}{17}} d x\)

= \(\sqrt{17} \cdot\left\{\frac{x}{2} \sqrt{x^2-\frac{11}{17}}-\frac{11}{34} \log \left|x+\sqrt{x^2-\frac{11}{17}}\right|\right\}+C\)

= \(\frac{x}{2} \sqrt{17 x^2-11}-\frac{11 \sqrt{17}}{34} \log \left|x+\sqrt{x^2-\frac{11}{17}}\right|+C .\)

Example 3 Evaluate \(\int \sqrt{16 x^2+25} d x\)

Solution

\(\int \sqrt{16 x^2+25} d x\)

We know that \(\int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|+C\)

∴ \(\int \sqrt{16 x^2+25} d x=4 \int\left\{\sqrt{x^2+\frac{25}{16}}\right\} d x=4 \int\left\{\sqrt{x^2+\left(\frac{5}{4}\right)^2}\right\} d x\)

= \(4 \cdot\left\{\frac{x}{2} \sqrt{x^2+\frac{25}{16}}+\frac{25}{32} \log \left|x+\sqrt{x^2+\frac{25}{16}}\right|\right\}+C\)

= \(\frac{x}{2} \cdot \sqrt{16 x^2+25}+\frac{25}{8} \log \left|x+\sqrt{x^2+\frac{25}{16}}\right|+C\)

WBCHSE Maths Chapter Special Integrals 

Example 4 Evaluate \(\int \sqrt{\frac{16+(\log x)^2}{x}} d x .\)

Solution

\(\int \sqrt{\frac{16+(\log x)^2}{x}} d x .\)

Putting log x = t and \(\frac{1}{x}\)dx = dt, we get

I = \(\int \sqrt{16+t^2} d t=\int \sqrt{4^2+t^2} d t\)

= \(\frac{t}{2} \sqrt{16+t^2}+\frac{16}{2} \log \left|t+\sqrt{16+t^2}\right|+C\)

= \(\frac{1}{2} \log x \cdot \sqrt{16+(\log x)^2}+8 \log \left|\log x+\sqrt{16+(\log x)^2}\right|+C\)

Example 5 Evaluate \(\int x \sqrt{x^4+1} d x\)

Solution

\(\int x \sqrt{x^4+1} d x\)

Putting x2 = t and x dx = \(\frac{1}{x}\)dt, we get

I = \(\frac{1}{2} \int \sqrt{t^2+1} d t\)

= \(\frac{1}{2} \cdot\left[\frac{t}{2} \sqrt{t^2+1}+\frac{1}{2} \log \left|t+\sqrt{t^2+1}\right|\right]+C\)

= \(\frac{x^2}{4} \sqrt{x^4+1}+\frac{1}{4} \log \left|x^2+\sqrt{x^4+1}\right|+C\)

Example 6 Evaluate \(\int e^x \sqrt{e^{2 x}+4} d x\)

Solution

\(\int e^x \sqrt{e^{2 x}+4} d x\)

Putting ex = t and exdx = dt, we get

I = \(\int \sqrt{t^2+4} d x\)

= \(\frac{t}{2} \cdot \sqrt{t^2+4}+\frac{4}{2} \log \left|t+\sqrt{t^2+4}\right|+C\)

= \(\frac{1}{2} e^x \sqrt{e^{2 x}+4}+2 \log \left|e^x+\sqrt{e^{2 x}+4}\right|+C\)

\(\int e^x \sqrt{e^{2 x}+4} d x\) = \(\frac{1}{2} e^x \sqrt{e^{2 x}+4}+2 \log \left|e^x+\sqrt{e^{2 x}+4}\right|+C\)

Example 7 Evaluate \(\int \cos x \sqrt{4-\sin ^2 x} d x .\)

Solution

\(\int \cos x \sqrt{4-\sin ^2 x} d x .\)

Putting sin x = t and cos x dx = dt, we get

I = \(\int \sqrt{4-t^2} d t\)

= \(\frac{t}{2} \sqrt{4-t^2}+\frac{4}{2} \sin ^{-1} \frac{t}{2}+C\)

= \(\frac{1}{2} \sin x \sqrt{4-\sin ^2 x}+2 \sin ^{-1}\left(\frac{1}{2} \sin x\right)+C .\)

Integrals of the form \(\int \sqrt{\left(a x^2+b x+c\right)} d x\)

Method Express (ax2 + bx + c) as a[(x+α)2 ± β2] and obtain an integral which can be evaluated easily.

Example 8 Evaluate \(\int \sqrt{x^2+3 x} d x .\)

Solution We have

\(\left(x^2+3 x\right)=\left\{x^2+3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\right\}=\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)

∴ \(I=\int \sqrt{x^2+3 x} d x\)

= \(\int \sqrt{\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2} d x=\int \sqrt{t^2-\left(\frac{3}{2}\right)^2} d t\), where (x + \(\frac{3}{2}\)) = t

= \(\frac{1}{2} t \sqrt{t^2-\frac{9}{4}}-\frac{9}{8} \log \left|t+\sqrt{t^2-\frac{9}{4}}\right|+C\)

{using \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|+C\)}

= \(\frac{1}{2}\left(x+\frac{3}{2}\right) \sqrt{x^2+3 x}-\frac{9}{8} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^2+3 x}\right|+C\)

WBCHSE Maths Chapter Special Integrals 

Example 9 Evaluate \(\int \sqrt{2 x^2+3 x+4} d x\)

Solution We have

\(\left(2 x^2+3 x+4\right)=2\left(x^2+\frac{3}{2} x+2\right)\)

= \(2 \cdot\left\{\left(x^2+\frac{3}{2} x+\frac{9}{16}\right)+\left(2-\frac{9}{16}\right)\right\}=2 \cdot\left\{\left(x+\frac{3}{4}\right)^2+\left(\frac{\sqrt{23}}{4}\right)^2\right\} .\)

∴ \(\sqrt{2 x^2+3 x+4}=\sqrt{2} \cdot \sqrt{\left(x+\frac{3}{4}\right)^2+\left(\frac{\sqrt{23}}{4}\right)^2}\)

⇒ \(\int \sqrt{2 x^2+3 x+4} d x=\sqrt{2} \cdot \int \sqrt{\left(x+\frac{3}{4}\right)^2+\left(\frac{\sqrt{23}}{4}\right)^2} d x\)

= \(\sqrt{2} \int \sqrt{t^2+\left(\frac{\sqrt{23}}{4}\right)^2} d t\), where (x + \(\frac{3}{2}\)) = t

= \(\sqrt{2} \cdot\left\{\frac{t}{2} \cdot \sqrt{t^2+\frac{23}{16}}+\frac{23}{32} \log \left|t+\sqrt{t^2+\frac{23}{16}}\right|\right\}+C\)

[∵ \(\int \sqrt{t^2+a^2} d t=\frac{t}{2} \sqrt{t^2+a^2}+\frac{a^2}{2} \log \left|t+\sqrt{t^2+a^2}\right|+C\)]

= \(\frac{\sqrt{2}}{2}\left(x+\frac{3}{4}\right) \sqrt{\left(x+\frac{3}{4}\right)^2+\frac{23}{16}}+\frac{23 \sqrt{2}}{32} \log \left|\left(x+\frac{3}{4}\right)+\sqrt{x^2+\frac{3}{2} x+2}\right|+C\)

= \(\frac{(4 x+3)}{4 \sqrt{2}} \cdot \sqrt{x^2+\frac{3}{2} x+2}+\frac{23 \sqrt{2}}{32} \log \left|\frac{(4 x+3)}{4}+\frac{\sqrt{2 x^2+3 x+4}}{\sqrt{2}}\right|+C\)

= \(\frac{(4 x+3) \sqrt{2 x^2+3 x+4}}{8}+\frac{23 \sqrt{2}}{32} \log \left|\frac{(4 x+3)}{4}+\frac{\sqrt{2 x^2+3 x+4}}{\sqrt{2}}\right|+C \text {. }\)

Example 10 Evaluate \(\int \sqrt{3-2 x-2 x^2} d x\)

Solution We have

\(\left(3-2 x-2 x^2\right)=2 \cdot\left\{\frac{3}{2}-x-x^2\right\}\)

= \(2 \cdot\left[\frac{3}{2}-\left(x^2+x+\frac{1}{4}\right)+\frac{1}{4}\right]\)

= \(\left[\frac{7}{4}-\left(x+\frac{1}{2}\right)^2\right]=2 \cdot\left\{\left(\frac{\sqrt{7}}{2}\right)^2-\left(x+\frac{1}{2}\right)^2\right\}\)

∴ \(\sqrt{3-2 x-2 x^2}=\sqrt{2} \cdot \sqrt{\left(\frac{\sqrt{7}}{2}\right)^2-\left(x+\frac{1}{2}\right)^2}\)

⇒ \(\int \sqrt{3-2 x-2 x^2} d x=\sqrt{2} \cdot \int \sqrt{\left(\frac{\sqrt{7}}{2}\right)^2-\left(x+\frac{1}{2}\right)^2} d x\)

= \(\sqrt{2} \cdot \int \sqrt{\left(\frac{\sqrt{7}}{2}\right)^2-t^2} d t\), where (x + \(\frac{1}{2}\)) = t and dx = dt

= \(\sqrt{2} \cdot\left\{\frac{t}{2} \cdot \sqrt{\frac{7}{4}-t^2}+\frac{7}{8} \sin ^{-1} \frac{t}{(\sqrt{7 / 2})}\right\}+C\)

[∵ \(\int \sqrt{a^2-t^2} d t=\frac{t}{2} \sqrt{a^2-t^2}+\frac{a^2}{2} \cdot \sin ^{-1} \frac{t}{a}+C\)]

= \(\sqrt{2} \cdot\left\{\frac{1}{2}\left(x+\frac{1}{2}\right) \sqrt{\frac{7}{4}-\left(x+\frac{1}{2}\right)^2}+\frac{7}{8} \sin ^{-1} \frac{\left(x+\frac{1}{2}\right)}{(\sqrt{7} / 2)}\right\}+C\)

= \(\sqrt{2}\left\{\frac{1}{4}(2 x+1) \cdot \sqrt{\frac{3}{2}-x-x^2}+\frac{7}{8} \sin ^{-1}\left(\frac{2 x+1}{\sqrt{7}}\right)\right\}+C\)

= \(\frac{1}{4}(2 x+1) \sqrt{3-2 x-2 x^2}+\frac{7}{4 \sqrt{2}} \sin ^{-1}\left(\frac{2 x+1}{\sqrt{7}}\right)+C \text {. }\)

Integrals of the form \(\int(p x+q) \sqrt{\left(a x^2+b x+c\right)} d x\)

Method Let (px+q) = \(A \cdot \frac{d}{d x}\left(a x^2+b x+c\right)+B\).

Find A and B.

Then, we get the integrand which is easily integrable.

Example 11 Evaluate \(\int(x+3) \sqrt{3-4 x-x^2} d x .\)

Solution

\(\int(x+3) \sqrt{3-4 x-x^2} d x .\)

Let (x+3) = \(A \cdot \frac{d}{d x}\left(3-4 x-x^2\right)+B ⇔ (x+3) = A(-4-2x) + B\).

Comparing coefficients of like powers of x, we get

-2A = 1 and -4A + B = 3 ⇔ A = \(\frac{1}{2}\) and B = 1.

∴ \((x+3)=-\frac{1}{2}(-4-2 x)+1\)

⇒ \(I=\int\left\{-\frac{1}{2}(-4-2 x)+1\right\} \sqrt{3-4 x-x^2} d x\)

= \(-\frac{1}{2} \int(-4-2 x) \sqrt{3-4 x-x^2} d x+\int \sqrt{3-4 x-x^2} d x\)

= \(-\frac{1}{2} \int \sqrt{t} d t+\int \sqrt{7-\left(4 x+x^2+4\right)} d x\), where 3-4x-x2 = t

= \(\left(-\frac{1}{2} \times t^{3 / 2} \times \frac{2}{3}\right)+\int \sqrt{(\sqrt{7})^2-(x+2)^2} d x\)

= \(-\frac{1}{3}\left(3-4 x-x^2\right)^{3 / 2}+\frac{1}{2}(x+2) \sqrt{3-4 x-x^2}+\frac{7}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C\)

[∵ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \cdot \sin ^{-1} \frac{x}{a}+C\)]

Example 12 Evaluate \(\int(3 x-2) \sqrt{x^2+x+1} d x .\)

Solution

\(\int(3 x-2) \sqrt{x^2+x+1} d x .\)

Let (3x-2) = \(A \cdot \frac{d}{d x}\left(x^2+x+1\right)+B\)

Then, (3x-2) = A(2x+1) + b.

Comparing coefficients of like powers of x, we get

2A = 3 and A + B = -2. So, A = \(\frac{3}{2}\) and B = \(-\frac{7}{2}\)

∴ \((3 x-2)=\frac{3}{2}(2 x+1)-\frac{7}{2} .\)

So, \(\int(3 x-2) \sqrt{x^2+x+1} d x\)

= \(\int\left\{\frac{3}{2}(2 x+1)-\frac{7}{2}\right\} \sqrt{x^2+x+1} d x\)

= \(\frac{3}{2} \int(2 x+1) \sqrt{x^2+x+1} d x-\frac{7}{2} \int \sqrt{x^2+x+1} d x\)

= \(\frac{3}{2} \int \sqrt{t} d t-\frac{7}{2} \int \sqrt{\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}} d x\), where x2 + x + 1 = t in the 1st integral

= \(t^{3 / 2}-\frac{7}{2} \int \sqrt{\left\{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right\}} d x\)

= \(\left(x^2+x+1\right)^{3 / 2}-\frac{7}{2} \cdot \int \sqrt{u^2+\left(\frac{\sqrt{3}}{2}\right)^2} d u\), Where u = (x + \frac{1}{2})

= \(\left(x^2+x+1\right)^{3 / 2}-\frac{7}{2}\left\{\frac{u}{2} \cdot \sqrt{u^2+\frac{3}{4}}+\frac{3}{8} \log \left|u+\sqrt{u^2+\frac{3}{4}}\right|\right\}+C\)

{∵ \(\int \sqrt{u^2+a^2} d u=\frac{u}{2} \sqrt{u^2+a^2}+\frac{a^2}{2} \log \left|u+\sqrt{u^2+a^2}\right|+C\)}

= \(\left(x^2+x+1\right)^{3 / 2}-\frac{7 u}{8} \sqrt{4 u^2+3}-\frac{21}{16} \log \left|u+\sqrt{u^2+\frac{3}{4}}\right|+C\)

= \(\left(x^2+x+1\right)^{3 / 2}-\frac{7}{8}\left(x+\frac{1}{2}\right) \sqrt{4\left(x+\frac{1}{2}\right)^2+3}-\frac{21}{16} \log \mid\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4} \mid}+C\)

= \(\left(x^2+x+1\right)^{3 / 2}-\frac{7(2 x+1)}{8} \sqrt{x^2+x+1}-\frac{21}{16} \log \left|\frac{(2 x+1)}{2}+\sqrt{x^2+x+1}\right|+C \text {. }\)

WBCHSE Maths Chapter Special Integrals 

Example 13 Evaluate \(\int x \sqrt{x+x^2} d x\)

Solution

\(\int x \sqrt{x+x^2} d x\)

Let x = \(A \cdot \frac{d}{d x}\left(x+x^2\right)+B\). Then,

x = A(1+2x) + B …(1)

Comparing coefficients of various powers of x, we get

(2A = 1 and A + B = 0) ⇒ [A = \(\frac{1}{2}\) and B = \(-\frac{1}{2}\)].

∴ \(x=\frac{1}{2}(1+2 x)-\frac{1}{2}\)

⇒ \(\int x \sqrt{x+x^2} d x\)

= \(\int\left\{\frac{1}{2}(1+2 x)-\frac{1}{2}\right\} \sqrt{x+x^2} d x\)

= \(\frac{1}{2} \int(1+2 x) \sqrt{x+x^2} d x-\frac{1}{2} \int \sqrt{x+x^2} d x\)

= \(\frac{1}{2} \int \sqrt{t} d t-\frac{1}{2} \int \sqrt{\left\{\left(x^2+x+\frac{1}{4}\right)-\frac{1}{4}\right\}} d x\), where (x+x2) = t in the first integral

= \(\frac{1}{2} \cdot \frac{t^{3 / 2}}{(3 / 2)}-\frac{1}{2} \int \sqrt{\left\{\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right\}} d x\)

= \(\frac{1}{3}\left(x+x^2\right)^{3 / 2}-\frac{1}{2} \cdot \int \sqrt{u^2-\left(\frac{1}{2}\right)^2} d u\), where (x + \(\frac{1}{2}\)) = u

= \(\frac{1}{3}\left(x+x^2\right)^{3 / 2}-\frac{1}{2}\left\{\frac{u}{2} \cdot \sqrt{u^2-\frac{1}{4}}-\frac{1}{8} \log \left|u+\sqrt{u^2-\frac{1}{4}}\right|\right\}+C\)

{∵ \(\int \sqrt{x^2-a^2} d x=\frac{x}{2} \sqrt{x^2-a^2}-\frac{a^2}{2} \log \left|x+\sqrt{x^2-a^2}\right|+C\)}

= \(\frac{1}{3}\left(x+x^2\right)^{3 / 2}-\frac{1}{4}\left(x+\frac{1}{2}\right) \sqrt{\left(x+\frac{1}{2}\right)^2-\frac{1}{4}}+\frac{1}{16} \log \mid\left(x+\frac{1}{2}\right) \sqrt{\left(x+\frac{1}{2}\right)^2-\frac{1}{4} \mid}+C\)

= \(\frac{1}{3}\left(x+x^2\right)^{3 / 2}-\frac{1}{8}(2 x+1) \sqrt{x+x^2}+\frac{1}{16} \log \left|\frac{(2 x+1)}{2} \cdot \sqrt{x+x^2}\right|+C\)

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