Question

If $f\left(x\right)={\int }_0^{\sin x}\cos {}^{-1}tdt+{\int }_0^{\cos x}\sin {}^{-1}tdt,0<x<\frac{\pi }{2}$

then$f\left(\frac{\pi }{4}\right)$ is?

• A. $\frac{\pi }{\sqrt{2}}$
• B. $1+\frac{\pi }{2\sqrt{2}}$
• C. $1$
• D. none of these

Answer 1

Answer: (B)

$1+\frac{\pi }{2\sqrt{2}}$

$f\left(x\right)={\int }_0^{\sin x}{\cos }^{-1}tdt+{\int }_0^{\cos x}{\sin }^{-1}tdt$
$={\left[{\cos }^{-1}tt\right]}_0^{\sin x}+{\int }_0^{\sin x}\frac{1}{\sqrt{1-x^2}}tdt$
$+{[{\sin }^{-1}t.t]}_0^{\cos x}-{\int }_0^{\cos x}\frac{1}{\sqrt{1-x^2}}.t.dt$
$=\left({\cos }^{-1}\sin x\right)\sin x-0+{[-\sqrt{1-t^2}]}_0^{\cos x}+\left({\sin }^{-1}\left(\cos x\right)\right)\cos x-0-{[-\sqrt{1-t^2}]}_0^{\cos x}=\left({\cos }^{-1}\sin x\right)\sin x+\left({\sin }^{-1}\cos x\right)\cos x-\sqrt{1-{\sin }^{2}x}+1+\sqrt{1-{\cos }^{2}x}-1=\sin x\left({\cos }^{-1}\sin x\right)+\cos x\left({\sin }^{-1}\cos x\right)-\cos x+\sin x$
$\therefore f\left(\frac{\pi }{4}\right)=1+\frac{\pi }{2\sqrt{2}}$