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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals5 Marks
Question
Integrate the function $\frac{e^{a \tan ^{-1} x}}{\left(1+x^2\right)^{3 / 2}}$ with respect to $x$.
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Answer

$\text { Let }$
$I=\int \frac{e^{a \tan ^{-1} x}}{\left(1+x^2\right)^{3 / 2}} d x$
$\text { Putting } \tan ^{-1} x=t$
$\therefore x=\tan t$
$\therefore \frac{1}{1+x^2} d x=d t$
$I=\int \frac{e^{a \tan ^{-1} x}}{\sqrt{1+x^2}\left(1+x^2\right)} d x$
$=\int \frac{e^{a t} d t}{\sqrt{1+\tan ^2 t}}=\int \frac{e^{a t} d t}{\sec t}$
$I=\iint_{II}^{a t} \underset{I}{\cos t} d t$
$\text { (using ILATE) }$
$I=\cos t \cdot \frac{e^{a t}}{a}-\int \frac{(-\sin t) \cdot e^{a t}}{a} d t$
$=\frac{1}{a} e^{a t} \cos t+\frac{1}{a} \int_{II} e^{a t} \sin _{I} t d t$
$=\frac{1}{a} e^{a t} \cos t+ \frac{1}{a}\left(\frac{\sin t \cdot e^{a t}}{a}-\int \cos t \cdot \frac{e^{a t}}{a} d t\right)$
$=\frac{1}{a} e^{a t} \cos t+\frac{1}{a^2} e^{a t} \sin t-\frac{1}{a^2} \int e^{a t} \cos t d t$
$=\frac{1}{a^2} e^{a t}(a \cos t+\sin t)-\frac{1}{a^2} I$
$\left(1+\frac{1}{a^2}\right) I =\frac{e^{a t}}{a^2}(a \cos t+\sin t)+ C$
$I =\frac{e^{a t}}{a^2} \times \frac{a^2}{\left(a^2+1\right)}(a \cos t+\sin t)+ C$
$= \frac{e^{a t}}{\left(a^2+1\right)}(a \cos t+\sin t)+ C$
$\therefore \tan -1 x=t \Rightarrow x=\tan t$
$= \frac{e^{a \tan ^{-1} x}}{a^2+1}\left(a \cdot \frac{1}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}\right)+ C$
$= \frac{e^{a \tan ^{-1} x}}{\left(1+a^2\right)} \cdot \frac{(a+x)}{\sqrt{1+x^2}}+ C Ans .$

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