Question

The value of $\left({\int }_0^{{\sin }^{2}x}{\sin }^{-1}\sqrt{t}dt\right)+\left({\int }_0^{{\cos }^{2}x}{\cos }^{-1}\sqrt{t}dt\right)$ is

• A. $\pi /2$
• B. $1$
• C. $\pi /4$
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A $\pi /2$
B $1$

We have
$I={\int }_0^{\sin x}{\sin }^{-1}\left(\sqrt{t}\right)dt+{\int }_0^{{\cos }^{2}x}{\cos }^{-1}\left(\sqrt{t}\right)dt$

$={\left[t{\sin }^{-1}\left(\sqrt{t}\right)\right]}_0^{{\sin }^{2}x}-{\int }_0^{{\sin }^{2}x}\frac{\sqrt{t}}{\sqrt[2]{21-t}}dt+{\left[t{\cos }^{-1}\left(\sqrt{t}\right)\right]}_0^{{\cos }^{2}x}-{\int }_0^{{\cos }^{2}x}\frac{\sqrt{t}}{\sqrt[2]{21-t}}dt$

$=x{\sin }^{2}x+{\int }_{{\sin }^{2}x}^0\frac{\sqrt{t}}{\sqrt[2]{21-t}}dt+x{\cos }^{2}x+{\int }_0^{{\cos }^{2}x}\frac{\sqrt{t}}{\sqrt[2]{21-t}}dt$

$=x({\sin }^{2}x+{\cos }^{2}x)+{\int }_{{\cos }^{2}x}^{{\cos }^{2}x}\frac{\sqrt{t}}{\sqrt[2]{21-t}}dt$
Substitute $={\sin }^2\theta$ and $dt=2\sin \theta \cos \theta d\theta ,$ we get,

$\int \frac{\sqrt{t}}{\sqrt[2]{21-t}}dt=\int \frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}2\sin \theta \cos \theta d\theta$

$=\int {\sin }^2\theta d\theta =\int \frac{1-\cos 2\theta }{2}d\theta$

$=\frac{\theta }{2}-\frac{\sin 2\theta }{4}$

Also, when $t={\sin }^{2}x,\theta =x$ and when$t={\cos }^{2}x,\theta =\frac{\pi }{2}-x$

$\therefore I=x+{[\frac{\theta }{2}-\frac{\sin 2\theta }{4}]}_x^{\frac{\pi }{2}-x}$

$=x+(\frac{\pi }{4}-\frac{x}{2}-\frac{\sin 2x}{4})-(\frac{x}{2}-\frac{\sin 2x}{4})$

$=x+\frac{\pi }{2}-x=\frac{\pi }{4}$