Question

If $\int si{n}^{-1}xco{s}^{-1}xdx=f^{\prime }\left(x\right)$
$\left[\frac{\pi }{2}x-x{f}^{\prime }\left(x\right)-2\sqrt{1-x^2}\right]\frac{\pi }{2}\sqrt{1-x^2}+2x+C$, then

• A. $f\left(x\right)=sinx$
• B. $f\left(x\right)=cosx$
• C. $f\left(x\right)=tanx$
• D. None of these

Answer 1

Answer: (A)

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