Question

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

Answer 1

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

It is known that, ${\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}=\frac{\pi }{2}$

$\Rightarrow I=\int \frac{(\frac{\pi }{2}-{\cos }^{-1}\sqrt{x})-{\cos }^{-1}\sqrt{x}}{\frac{\pi }{2}}dx$
$=\frac{2}{\pi }\int (\frac{\pi }{2}-2{\cos }^{-1}\sqrt{x})dx$
$=\frac{2}{\pi }\cdot \frac{\pi }{2}\int 1\cdot dx-\frac{4}{\pi }\int {\cos }^{-1}\sqrt{x}dx$
$=x-\frac{4}{\pi }\int {\cos }^{-1}\sqrt{x}dx$ ............. (1)
Let $I_1=\int {\cos }^{-1}\sqrt{x}dx$
Also, let $\sqrt{x}=t\Rightarrow dx=2tdt$
$\Rightarrow I_1=2\int {\cos }^{-1}t\cdot tdt$
$=2[{\cos }^{-1}t\cdot \frac{t^2}{2}-\int \frac{-1}{\sqrt{1-t^2}}\cdot \frac{t^2}{2}dt]$
$=t^2{\cos }^{-1}t+\int \frac{t^2}{\sqrt{1-t^2}}dt$
$=t^2{\cos }^{-1}t-\int \frac{1-t^2-1}{\sqrt{1-t^2}}dt$
$=t^2{\cos }^{-1}t-\int \sqrt{1-t^2}dt+\int \frac{1}{\sqrt{1-t^2}}dt$
$=t^2{\cos }^{-1}t-\frac{t}{2}\sqrt{1-t^2}-\frac{1}{2}{\sin }^{-1}t+{\sin }^{-1}t$
$=t^2{\cos }^{-1}t-\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t$

From equation (1), we obtain

$I=x-\frac{4}{\pi }[t^2{\cos }^{-1}t-\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t]$
$=x-\frac{4}{\pi }[x{\cos }^{-1}\sqrt{x}-\frac{\sqrt{x}}{2}\sqrt{1-x}+\frac{1}{2}{\sin }^{-1}\sqrt{x}]$
$=x-\frac{4}{\pi }[x(\frac{\pi }{2}-{\sin }^{-1}\sqrt{x})-\frac{\sqrt{x-x^2}}{2}+\frac{1}{2}{\sin }^{-1}\sqrt{x}]$
$=x-2x+\frac{4x}{\pi }{\sin }^{-1}\sqrt{x}+\frac{2}{\pi }\sqrt{x-x^2}-\frac{2}{\pi }{\sin }^{-1}\sqrt{x}$

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

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