Question

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

Answer 1

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

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

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

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

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

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

$=\frac{4{\sin }^{-1}\sqrt{x}}{\pi }-1$

Integrate w.r.t to $x$

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

$=\int (\frac{4{\sin }^{-1}\sqrt{x}}{\pi }-1)dx$

$=\frac{4}{\pi }\int {\sin }^{-1}\sqrt{x}dx-x+C_1$

Let $I_1=\int {\sin }^{-1}\sqrt{x}dx$

Let $\sqrt{x}=t\Rightarrow x=t^2$

$\Rightarrow dx=2tdt$

$\therefore I_1=\int {\sin }^{-1}t.2tdt=2\int {\sin }^{-1}t.tdt$

Consider $\int {\sin }^{-1}t.tdt$ using integration by parts,

Let $u={\sin }^{-1}t\Rightarrow du=\frac{1}{\sqrt{1-t^2}dt}$ and $dv=tdt\Rightarrow v=\frac{t^2}{2}$

$\int {\sin }^{-1}t.tdt={\sin }^{-1}t\frac{t^2}{2}-\int \frac{t^2}{2}\frac{1}{\sqrt{1-t^2}}dt$

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

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

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

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

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

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

$=t^2{\sin }^{-1}t+\frac{t\sqrt{1-t^2}}{2}-\frac{1}{2}{\sin }^{-1}t$

$\therefore I_1=t^2{\sin }^{-1}t+\frac{t\sqrt{1-t^2}}{2}-\frac{1}{2}{\sin }^{-1}t$

Put $t=\sqrt{x}$

$I_1={\left(\sqrt{x}\right)}^2{\sin }^{-1}\sqrt{x}+\frac{\sqrt{x}\sqrt{1-{\left(\sqrt{x}\right)}^2}}{2}-\frac{1}{2}{\sin }^{-1}\sqrt{x}$

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

Hence $I=\frac{4}{\pi }{I}_1-x+C_1$

$=\frac{4}{\pi }({\left(\sqrt{x}\right)}^2{\sin }^{-1}\sqrt{x}+\frac{\sqrt{x}\sqrt{1-{\left(\sqrt{x}\right)}^2}}{2}-\frac{1}{2}{\sin }^{-1}\sqrt{x})-x+C_1$

$=\frac{4}{\pi }(x{\sin }^{-1}\sqrt{x}+\frac{\sqrt{x-x^2}}{2}-\frac{1}{2}{\sin }^{-1}\sqrt{x})-x+C_1$

$={\sin }^{-1}\sqrt{x}[\frac{4x}{\pi }-\frac{2}{\pi }]+\frac{2\sqrt{x-x^2}}{\pi }-x+C_1$

$={\sin }^{-1}\sqrt{x}\left[\frac{2(2x-1)}{\pi }\right]+\frac{2\sqrt{x-x^2}}{\pi }-x+C_1$