_0^1 ^ -1 (2 x 1+x^2 ) d x= - Vidyadip

MCQ

$\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=$

✓
$\frac{\pi}{2}-2 \log \sqrt{2}$
B
$\frac{\pi}{2}+2 \log \sqrt{2}$
C
$\frac{\pi}{4}-\log \sqrt{2}$
D
$\frac{\pi}{4}+\log \sqrt{2}$

✓

Answer

Correct option: A.

$\frac{\pi}{2}-2 \log \sqrt{2}$

(A)
Put $x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$
When $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{4}$
$\therefore \quad \int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=\int_0^{\frac{\pi}{4}} \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \sec ^2 \theta d \theta$
$\begin{array}{l}=2 \int_0^{\frac{\pi}{4}} \theta \sec ^2 \theta d \theta \\ =2[\theta \tan \theta]_0^{\pi / 4}-2 \int_0^{\frac{\pi}{4}} \tan \theta d \theta \\ =\frac{\pi}{2}+2[\log \cos x]_0^{\pi / 4} \\ =\frac{\pi}{2}-2 \log \sqrt{2}\end{array}$

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