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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Integrate the function $\text { tan }^{-1} \sqrt{\frac{1-x}{1+x}}$
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Answer

Let $I=\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$
Let x = cos $\theta$ $\Rightarrow$ dx = -sin $\theta$ d$\theta$
$\Rightarrow \theta=\cos ^{-1} x$ 
$\Rightarrow \mathrm{I}=\int \tan ^{-1} \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}} \mathrm{d} \mathrm{x}$ = $\int \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-\sin \theta) d \theta$ 
= $-\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)}}(\sin \theta) d \theta$ 
= $-\int \tan ^{-1} \sqrt{\tan ^{2}\left(\frac{\theta}{2}\right)}(\sin \theta) d \theta$ 
= $-\int \tan ^{-1} \tan \frac{\theta}{2} \cdot(\sin \theta) d \theta$ 
= $-\frac{1}{2} \int \theta \cdot(\sin \theta) d \theta$ 
As we know, $\int \mathrm{u} . \mathrm{v} \mathrm{dx}=\mathrm{u} . \int \mathrm{v} \mathrm{dx}-\int \frac{\mathrm{d}}{\mathrm{dx}}(u) \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$ 
 $\Rightarrow -\frac{1}{2} \int \theta \cdot(\sin \theta) d \theta$ = $-\frac{1}{2}\left[\theta . \int \sin \theta \mathrm{d} \theta-\int \frac{\mathrm{d}}{d \theta}{(\theta)} \cdot\left\{\int \sin \theta \mathrm{d} \theta\right\} \mathrm{d} \theta\right]$
= $-\frac{1}{2}\left[\theta \cdot(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta\right]$ 
= $-\frac{1}{2}[-\theta \cdot \cos \theta+\sin \theta]$ 
= $\frac{1}{2} \theta \cdot \cos \theta-\frac{1}{2} \sqrt{\left(1-\cos ^{2} \theta\right.})$ 
= $\frac{1}{2} \cos ^{-1} x \cdot x-\frac{1}{2} \sqrt{\left(1-x^{2}\right.}+C$ 
= $\frac{1}{2}\left(x . \cos ^{-1} x-\sqrt{\left(1-x^{2}\right.}\right)+C$

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