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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals5 Marks
Question
Evaluate the integral $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$ using substitution.
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Answer

Given: $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{x^{2}+1}\right) d x$
Let $x = \tan \theta$
$\Rightarrow d x = \sec^2 \theta d \theta$
When, $x = 0, \theta= 0$ and when $x = 1, \theta = \frac{\pi}{4}$
Let $I=\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{x^{2}+1}\right) d x$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \sin ^{-1}\left(\frac{2 \tan \theta}{\tan ^{2} \theta+1}\right) \sec ^{2} \theta \mathrm{d} \theta$
$= \int_{0}^{\frac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^{2} \theta d \theta$
$= \int_{0}^{\frac{\pi}{4}} 2 \theta \sec ^{2} \theta d \theta$
$= 2 \int_{0}^{\frac{\pi}{4}} \theta \sec ^{2} \theta d \theta$
By applying product rule as,
$\int \mathrm{u} \cdot \mathrm{v} \mathrm{d} \mathrm{x}=\mathrm{u} \cdot \int \mathrm{vdx}-\int \frac{\mathrm{du}}{\mathrm{dx}} \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=2\left[\theta \int \sec ^{2} \theta \mathrm{d} \theta-\int \frac{\mathrm{d}}{\mathrm{d} \theta} \theta \cdot\left\{\int \sec ^{2} \theta \mathrm{d} \theta\right\} \mathrm{d} \theta\right]_{0}^{\frac{\pi}{4}}$
$= 2\left[\theta \tan \theta-\int 1 . \tan \theta d \theta\right]_{0}^{\frac{\pi}{4}}$
$= 2[\theta \tan \theta-\log |\sec \theta|]_{0}^{\frac{\pi}{4}}$
$= 2\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\log \left|\sec \frac{\pi}{4}\right|-0+\log |\sec \theta|\right]$
$= 2\left[\frac{\pi}{4}-\frac{1}{2} \log (2)\right.]$
$= \frac{\pi}{2}-\log (2)$

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