Question

For $0 \leq x \leq \frac{\pi}{2}$, the value of $\int\limits_{0}^{\sin ^{2} x} \sin ^{-1}(\sqrt{t}) d t+\int\limits_{0}^{\cos ^{2} x} \cos ^{-1}(\sqrt{t}) d t$ equals :

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

Answer 1

Answer: (A)

Consider
$\int\limits_{0}^{\sin ^{2} x} \sin ^{-1}(\sqrt{t}) d t+\int\limits_{0}^{\cos ^{2} x} \cos ^{-1}(\sqrt{t}) d t$
Let $I=f(x)$ after in tegrating and putting the limits.
$f'(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'(x)=0 $
$\Rightarrow f(x)=C $ (constant)
Now, we find $f(x)$ at $x=\frac{\pi}{4}$
$\therefore I=\int\limits_{0}^{1 / 2} \sin ^{-1} \sqrt{t} d t+\int\limits_{0}^{1 / 2} \cos ^{-1} \sqrt{t} d t $
$=\int\limits_{0}^{1 / 2}\left(\sin ^{-1} \sqrt{t}+\cos ^{-1} \sqrt{t}\right) d t $
$=\int\limits_{0}^{1 / 2} \frac{\pi}{2} d t=\frac{\pi}{4}=C$
$\therefore f(x)=\frac{\pi}{4}$
$\therefore $ Required integration $=\frac{\pi}{4}$