Question

Integrate: $\frac{{\sin }^{-1}\sqrt{x}-{\cos }^{-1}\sqrt{x}}{{\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}},x\in [0,1]$

Answer 1

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

$=\int \frac{{\sin }^{-1}\sqrt{x}-{\cos }^{-1}\sqrt{x}}{\frac{\pi }{2}}dx$

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

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

$=\frac{4}{\pi }\int {\sin }^{-1}\sqrt{x}-\frac{2}{\pi }\times \frac{\pi }{2}.x$

$=I_1-x+C$ $\to$ (1)

$I_1=\frac{4}{\pi }\int {\sin }^{-1}\sqrt{x}.dx$ $\sqrt{x}=t$

$=\frac{4}{\pi }\int {\sin }^{-1}t\times 2t.dt$ $x=t^{2}dx=2t.dt$

$=\frac{8}{\pi }\int \frac{t^2}{2}{\sin }^{-1}t+\int \sqrt{1-t^2}\frac{dt}{2}-\int \frac{1}{\sqrt{1-t^2}}\frac{dt}{2}$

$=\frac{8}{\pi }[\frac{t^2}{2}{\sin }^{-1}t+\frac{t}{4}\sqrt{1-t^2}+\frac{1}{4}{\sin }^{-1}t-\frac{-1{\sin }^{-1}t}{2}]+C$

$\frac{x}{2}{\sin }^{-1}\sqrt{x}+\frac{\sqrt{x}}{4}\sqrt{1-x}+\frac{1}{4}{\sin }^{-1}\sqrt{x}\frac{-1}{2}{\sin }^{-1}\sqrt{x}$

from (1)

$=\frac{8}{\pi }\times \frac{1}{4}\left[\right(2x-1){\sin }^{-1}\sqrt{x}+\sqrt{x}(\sqrt{1-x}\left)\right]-x+C$

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