Question

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

Answer 1

Consider the following Integral.

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

Using formula:

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

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

$=\frac{2}{\pi }((x{\sin }^{-1}\sqrt{x}-\underset{}{\overset{}{\int }}\frac{1}{\sqrt{1-x^2}}xdx)-(x{\sin }^{-1}\sqrt{x}+\underset{}{\overset{}{\int }}\frac{1}{\sqrt{1-x^2}}xdx))$

Let , $t=1-x^2$ differentiate w.r.t x, we get.

$\frac{dt}{2x}=dx$

$=\frac{2}{\pi }((x{\sin }^{-1}\sqrt{x}-\underset{}{\overset{}{\int }}\frac{1}{\sqrt{t}}xdt)-(x{\sin }^{-1}\sqrt{x}+\underset{}{\overset{}{\int }}\frac{1}{\sqrt{t}}xdt))$

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

$I=\frac{2}{\pi }((x{\sin }^{-1}\sqrt{x}-2\sqrt{1-x^2})-(x{\sin }^{-1}\sqrt{x}+2\sqrt{1-x^2}))+C$

Hence, this is the correct answer.