<s>[INST] Evaluate $\int \sin^{-1}\frac{2x}{1+x^2}dx$ $\int \sin^{-1}\dfrac{2x}{1+x^2}dx$
My attempt is as follows:-
$$x=\tan\theta$$
$$dx=\sec^2\theta d\theta$$
$$\int \sin^{-1}(\sin2\theta) \cdot\sec^2\theta d\theta$$
So here should we make cases on the basis of values of $\theta$ or can we write $\sin^{-1}(\sin2\theta)$ as $2\theta$?
 [/INST] Let $ \displaystyle x=\tan \frac{\theta}{2} \text{ then }\displaystyle  d x=\frac{1}{2} \sec ^{2} \frac{\theta}{2} d \theta$ and
$\displaystyle \frac{2 x}{1+x^{2}}=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2\frac{\theta }{2}}=\sin \theta .$
$\therefore \displaystyle \int \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d \displaystyle  x=\int \theta \cdot \frac{1}{2} \sec ^{2} \frac{\theta}{2} d \theta$
$= \displaystyle \int \theta d\left(\tan \frac{\theta}{2}\right)$
$=\displaystyle \theta \tan \frac{\theta}{2}-\int \tan \frac{\theta}{2} d \theta$
$\displaystyle =x \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-2 \ln \left|\sec \frac{\theta}{2}\right| +C$
$=\displaystyle  x \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-\ln \left(1+x^{2}\right)+C$
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