Question

If ${\int }_0^{\pi /2}arctan\left(sinx\right)dx+{\int }_0^{\pi /4}arcsin\left(tanx\right)dx=\frac{{\pi }^2}{k}$,

then $k$ is divisible by?

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

Answer: (A), (C)

The correct options are
A $8$
C $4$

Given ${\int }_0^{\pi /2}ta{n}^{-1}\left(sinx\right)dx+{\int }_0^{\pi /4}si{n}^{-1}\left(tanx\right)dx$
Let $ta{n}^{-1}\left(sinx\right)=f\left(x\right)=y$ ....(1)
$\Rightarrow \sin x=\tan y$
$\Rightarrow x={\sin }^{-1}\left(\tan y\right)$
$\Rightarrow f^{-1}y={\sin }^{-1}\left(\tan y\right)$
$\Rightarrow f^{-1}\left(x\right)=si{n}^{-1}\left(tanx\right)$
So, ${\int }_0^{\pi /2}ta{n}^{-1}\left(sinx\right)dx+{\int }_0^{\pi /4}si{n}^{-1}\left(tanx\right)dx$
$={\int }_0^{\pi /2}f\left(x\right)dx+{\int }_0^{\pi /4}{f}^{-1}\left(x\right)dx$

$=\left[xf\right(x){]}_0^{\pi /2}$

$=\frac{\pi }{2}f\left(\frac{\pi }{2}\right)$
Now , by eqn (1),
$f\left(\frac{\pi }{2}\right)={\tan }^{-1}\sin \left(\frac{\pi }{2}\right)={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$
$\Rightarrow I_1+I_2=\frac{\pi }{2}\times \frac{\pi }{4}=\frac{{\pi }^2}{8}$
So, on comparing with given eqn
$k=8$
which is divisible by $4,8$