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

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Aicharya · Accepted Answer

$I=\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$ When, $x=\tan \theta \text {, }$ $\frac{2 x}{1+x^2}=\frac{2 \tan \theta}{1+\tan ^2 \theta}=\sin 2 \theta$ Also, $\quad d x=\sec ^2 \theta d \theta$ and, When $x=0, \theta=0$ When $x=1, \theta=\frac{\pi}{4}$ So, $\begin{aligned} I & =\int_0^{\pi / 4} \sin ^{-1}(\sin 2 \theta) \cdot \sec ^2 \theta \cdot d \theta \ & =\int_0^{\pi / 4} 2 \theta \cdot \sec ^2 \theta d \theta=2 \int_0^{\pi / 4} \theta \cdot \sec ^2 \theta \cdot d \theta \end{aligned}$ By parts we get, $I=2[\theta, \tan \theta-\log \sec \theta]_0^{\pi / 4} \Rightarrow I=\frac{\pi}{2}-\log 2$

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

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