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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
Evaluate: $\int_0^1\left(\frac{1}{1+x^2}\right) \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$
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Answer

Let $I =\int_0^1\left(\frac{1}{1+x^2}\right) \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$
Put $x=\tan \theta$
$\therefore d x=\sec ^2 \theta d \theta$
When $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{4}$
$ \therefore I=\int_0^{\frac{\pi}{4}}\left(\frac{1}{1+\tan ^2 \theta}\right) \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \sec ^2 \theta d \theta$
$=\int_0^{\frac{\pi}{4}}\left(\frac{1}{\sec ^2 \theta}\right) \sin ^{-1}(\sin 2 \theta) \sec ^2 \theta d \theta$
$=\int_0^{\frac{\pi}{4}} 2 \theta d \theta$
$=2\left[\frac{\theta^2}{2}\right]_0^{\frac{\pi}{4}}$
$=\left(\frac{\pi}{4}\right)^2-0$
$\therefore I=\frac{\pi^2}{16} $

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