Question

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

• A. $\frac{1}{\pi }[\sqrt{x-x^2}-(1-2x\left){\cos }^{-1}\sqrt{x}\right]-x+c$
• B. $\frac{2}{\pi }[\sqrt{x-x^2}-(1-2x\left){\sin }^{-1}\sqrt{x}\right]-x+c$
• C. $\frac{2}{\pi }[\sqrt{x-x^2}-{\sin }^{-1}\sqrt{x}]-x+c$
• D. $\frac{2}{\pi }[\sqrt{x-x^2}-(1-2x\left){\sin }^{-1}x\right]-x+c$

Answer 1

The correct option is B $\frac{2}{\pi }[\sqrt{x-x^2}-(1-2x\left){\sin }^{-1}\sqrt{x}\right]-x+c$

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

We know that
${\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}=\pi /2...\left(1\right)$
Also, ${\cos }^{-1}\sqrt{x}=\pi /2-{\sin }^{-1}\sqrt{x}...\left(2\right)$

Using equations (1) and (2), we get
$I=\frac{{\sin }^{-1}\sqrt{x}-(\pi /2-{\sin }^{-1}\sqrt{x})}{\pi /2}dx$
$=\frac{2}{\pi }\int (2{\sin }^{-1}\sqrt{x}-\pi /2)dx$
$=\frac{4}{\pi }\int \left({\sin }^{-1}\sqrt{x}\right)-\int 1dx$

Let $x={\sin }^2\theta \Rightarrow dx=2\sin \theta \cos \theta d\theta$
$I=\frac{4}{\pi }\int {\sin }^{-1}\left(\sin \theta \right)2\sin \theta \cos \theta d\theta -x+c$
$=\frac{4}{\pi }\int \theta \sin 2\theta d\theta -x+c$
$=\frac{4}{\pi }[\frac{-\theta \cos 2\theta }{2}+\int 1\times \frac{\cos 2\theta }{2}d\theta ]-x+c$
[Integrating by parts]
$=\frac{4}{\pi }[\frac{-\theta \cos 2\theta }{2}+\frac{\sin 2\theta }{4}]-x+c$
$=\frac{2}{\pi }[\sqrt{x-x^2}-(1-2x\left){\sin }^{-1}\sqrt{x}\right]-x+c$