Question

If $\int {\sin }^{-1}x{\cos }^{-1}xdx=f^{-1}\left(x\right)[\frac{\pi }{2}x-x{f}^{-1}\left(x\right)-2\sqrt{1-x^2}]\frac{\pi }{2}\sqrt{1-x^2}+2x+C$, then

• A. $f\left(x\right)=\sin x$
• B. $f\left(x\right)=\cos x$
• C. $f\left(x\right)=\tan x$
• D. None of these

Answer 1

Answer: (A)

$f\left(x\right)=\sin x$

Let $I=\int {\sin }^{-1}x{\cos }^{-1}xdx$
$=\int {\sin }^{-1}x(\frac{\pi }{2}-{\sin }^{-1}x)dx$
$=\frac{\pi }{2}\int {\sin }^{-1}xdx-\int {\left({\sin }^{-1}x\right)}^{2}dx$
$=\frac{\pi }{2}[x{\sin }^{-1}x+\sqrt{1-x^2}]-[x\cdot {\left({\sin }^{-1}x\right)}^2+2{\sin }^{-1}x\sqrt{1-x^2}-2x]+C$ [using integration by parts]
$={\sin }^{-1}x(\frac{\pi }{2}x-x{\sin }^{-1}x-2\sqrt{1-x^2})+\frac{\pi }{2}\sqrt{1-x^2}+2x+C$
$\therefore f^{-1}\left(x\right)={\sin }^{-1}x\Rightarrow f\left(x\right)=\sin x$