Question

Solve $\int \frac{{\sin }^{-1}\sqrt{x}-{\cos }^{-1}\sqrt{x}}{{\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}}dx$

Answer 1

$\int \frac{{\sin }^{-1}\sqrt{x}-{\cos }^{-1}\sqrt{x}}{{\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}}dx$

We know that

${\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}=\frac{\pi }{2}$ .......$\left(1\right)$

Also, ${\cos }^{-1}\sqrt{x}=\frac{\pi }{2}-{\sin }^{-1}\sqrt{x}$ .....$\left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$ we get

$\Rightarrow \int \frac{{\sin }^{-1}\sqrt{x}-(\frac{\pi }{2}-{\sin }^{-1}\sqrt{x})}{\pi /2}dx$

$\Rightarrow \frac{2}{\pi }\int (2{\sin }^{-1}\sqrt{x}-\frac{\pi }{2})dx$

$\Rightarrow \frac{4}{\pi }\int {\sin }^{-1}\sqrt{x}dx-\int 1dx$

Let $x={\sin }^2\theta \Rightarrow dx=2\sin \theta \cos \theta d\theta$

$\Rightarrow \frac{4}{\pi }\int {\sin }^{-1}\left(\sin \theta \right)2\sin \theta \cos \theta d\theta -x+C$

$\Rightarrow \frac{4}{\pi }\int \theta \sin 2\theta d\theta -x+C$

$\Rightarrow \frac{4}{\pi }[\frac{-\theta \cos 2\theta }{2}+\int 1\times \frac{\cos 2\theta }{2}d\theta ]-x+C$

Integrate by parts

$\Rightarrow \frac{4}{\pi }[\frac{-\theta \cos 2\theta }{2}+\frac{\sin 2\theta }{4}]-x+C$

$\Rightarrow \frac{4}{4\times \pi }[-2{\sin }^{-1}\sqrt{x}(1-2x)+2\sqrt{x}\sqrt{1-x}]-x+C$

$\Rightarrow \frac{2}{\pi }[\sqrt{x-x^2}-(1-2x\left){\sin }^{-1}\sqrt{x}\right]-x+C$