Question

If $f\left(x\right)={\int }_0^{{\sin }^{2}x}{\sin }^{-1}\sqrt{t}dt+{\int }_0^{{\cos }^{2}x}{\cos }^{-1}\sqrt{t}dt,x\in [0,\frac{\pi }{2}]$, then $f\left(x\right)$ is equal to

• A. $f\left(x\right)=\frac{\pi }{6}$
• B. $f\left(x\right)=\frac{\pi }{8}$
• C. $f\left(x\right)=\frac{\pi }{4}$
• D. $f\left(x\right)=\frac{\pi }{2}$

Answer 1

The correct option is C $f\left(x\right)=\frac{\pi }{4}$

We have, $f\left(x\right)={\int }_0^{{\sin }^{2}x}{\sin }^{-1}\sqrt{tdt}+{\int }_0^{{\cos }^{2}x}{\cos }^{-1}\sqrt{tdt}$

Put $t={\sin }^2\theta$ in $1^{st}$ integral

so, $t\to 0\Rightarrow \theta \to 0$ and $t\to si{n}^{2}x\Rightarrow \theta \to x$

and $t={\cos }^2\varphi$ in the second integral, then

So, $t\to 0\Rightarrow \varphi \to \frac{\pi }{2}$ and $t\to si{n}^{2}x\Rightarrow \varphi \to x$

$f\left(x\right)={\int }_0^x\theta \sin 2\theta d\theta -{\int }_{\frac{\pi }{2}}^x\varphi \sin 2\varphi d\varphi$

$f\left(x\right)={\int }_0^x\theta \sin 2\theta d\theta +{\int }_x^{\frac{\pi }{2}}\theta \sin 2\theta d\theta$

$f\left(x\right)={\int }_0^{\frac{\pi }{2}}\theta \sin 2\theta d\theta =\frac{\pi }{4}$ (Using Integration by Parts).