Question

For $0\le x\le \frac{\pi }{2}$, the value of $\underset{0}{\overset{{\sin }^{2}x}{\int }}{\sin }^{-1}(\sqrt{t})dt+\underset{0}{\overset{{\cos }^{2}x}{\int }}{\cos }^{-1}(\sqrt{t})dt$ equals :

• A. $\frac{\pi }{4}$
• B. $0$
• C. $1$
• D. $-\frac{\pi }{4}$

Answer 1

Answer: (A)

Consider
$\underset{0}{\overset{{\sin }^{2}x}{\int }}{\sin }^{-1}(\sqrt{t})dt+\underset{0}{\overset{{\cos }^{2}x}{\int }}{\cos }^{-1}(\sqrt{t})dt$
Let $I=f(x)$ after in tegrating and putting the limits.
$f^{\prime }(x)={\sin }^{-1}\sqrt{{\sin }^{2}x}(2\sin x\cos x)-0$
$+{\cos }^{-1}\sqrt{{\cos }^{2}x}(-2\cos x\sin x)-0$
$\therefore f^{\prime }(x)=0$
$\Rightarrow f(x)=C$ (constant)
Now, we find $f(x)$ at $x=\frac{\pi }{4}$
$\therefore I=\underset{0}{\overset{1/2}{\int }}{\sin }^{-1}\sqrt{t}dt+\underset{0}{\overset{1/2}{\int }}{\cos }^{-1}\sqrt{t}dt$
$=\underset{0}{\overset{1/2}{\int }}\left({\sin }^{-1}\sqrt{t}+{\cos }^{-1}\sqrt{t}\right)dt$
$=\underset{0}{\overset{1/2}{\int }}\frac{\pi }{2}dt=\frac{\pi }{4}=C$
$\therefore f(x)=\frac{\pi }{4}$
$\therefore$ Required integration $=\frac{\pi }{4}$