Question

$\int \sin \left\{2{\tan }^{-1}\sqrt{\frac{3-x}{3+x}}\right\}dx=\frac{3}{2}{\cos }^{-1}\frac{x}{3}+\frac{3}{4}\sin \left(2{\cos }^{-1}\frac{x}{3}\right)$. If this is true enter 1, else enter 0.

Answer 1

Let $I=\int \sin \left\{2{\tan }^{-1}\sqrt{\frac{3-x}{3+x}}\right\}dx$
Put $x=3\cos \theta \Rightarrow dx=-3\sin \theta d\theta$
And using $\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}=\tan \frac{\theta }{2}$
We get
$I=-3\int \sin \theta \sin \theta d\theta =-\frac{3}{2}\int (1-\cos 2\theta )d\theta$
$=\frac{-3}{2}\{\theta -\frac{\sin 2\theta }{2}\}=\frac{-3}{2}{\cos }^{-1}\frac{x}{3}+\frac{3}{4}\sin \left(2{\cos }^{-1}\frac{x}{3}\right)$