Question

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

Answer 1

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

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

Substituting this, we get

$\Rightarrow \frac{2}{\pi }\int si{n}^{-1}x+co{s}^{-1}xdx$

$\Rightarrow \frac{2}{\pi }\int si{n}^{-1}xdx+\frac{2}{\pi }\int co{s}^{-1}xdx$

Use integration by parts, we get

$\Rightarrow \frac{2}{\pi }\int si{n}^{-1}x.1dx+\frac{2}{\pi }\int co{s}^{-1}x.1dx$

$\Rightarrow \frac{2}{\pi }({\sin }^{-1}x.x-\int \frac{x}{\sqrt{1-x^2}}dx)+\frac{2}{\pi }(co{s}^{-1}x.x+\int \frac{x}{\sqrt{1-x^2}}dx)$

$\Rightarrow \frac{2}{\pi }(x{\sin }^{-1}x-\frac{1}{2}\int \frac{2x}{\sqrt{1-x^2}}dx)+\frac{2}{\pi }(xco{s}^{-1}x+\frac{1}{2}\int \frac{2x}{\sqrt{1-x^2}}dx)$

Take $1-x^2=t\Rightarrow -2xdx=dt$

Substituting this we get

$\Rightarrow \frac{2}{\pi }(x{\sin }^{-1}x+\frac{1}{2}\int \frac{dt}{\sqrt{t}})+\frac{2}{\pi }(xco{s}^{-1}x-\frac{1}{2}\int \frac{dt}{\sqrt{t}})$

$\Rightarrow \frac{2}{\pi }(x{\sin }^{-1}x+\frac{1}{2}\sqrt{t}\times 2)+\frac{2}{\pi }(xco{s}^{-1}x-\frac{1}{2}\sqrt{t}\times 2)$

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

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