Question

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

• A. $\frac{4}{\pi }[x{\sin }^{-1}+\sqrt{1-x^2}]-x+c$
• B. $\log [{\sin }^{-1}+\sqrt{1-x^2}]-x+c$
• C. $\frac{4}{\pi }[x{\sin }^{-1}+\sqrt{1-x^2}]+c$
• D. $\frac{2}{\pi }[x{\sin }^{-1}x-x{\cos }^{-1}x+2\sqrt{1-x^2}]+c$

Answer 1

The correct option is D $\frac{2}{\pi }[x{\sin }^{-1}x-x{\cos }^{-1}x+2\sqrt{1-x^2}]+c$

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

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

$\frac{2}{\pi }\int {\sin }^{-1}x\int {\cos }^{-1}xdx$

$\Rightarrow \frac{2}{\pi }[x{\sin }^{-1}x+\sqrt{1-x^2}-x{\cos }^{-1}x+\sqrt{1-x^2}]+c$

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