For $0 \leq x \leq \frac{\pi}{2}$, the value of $ \int_0^{\sin ^2 x} \sin ^{-1}(\sqrt{t}) d t+\int_0^{\cos…
For $0 \leq x \leq \frac{\pi}{2}$, the value of
$
\int_0^{\sin ^2 x} \sin ^{-1}(\sqrt{t}) d t+\int_0^{\cos ^2 x} \cos ^{-1}(\sqrt{t}) d t \text { equals : }
$
$\frac{\pi}{4}$
0
1
$-\frac{\pi}{4}$
Solution
Consider
$
\int_0^{\sin ^2 x} \sin ^{-1}(\sqrt{t}) d t+\int_0^{\cos ^2 x} \cos ^{-1}(\sqrt{t}) d t
$
Let $\mathrm{I}=f(x)$ after integrating and putting the limits.
$
\begin{aligned}
&f^{\prime}(x)=\sin ^{-1} \sqrt{\sin ^2 x}(2 \sin x \cos x)-0 \\
&\quad+\cos ^{-1} \sqrt{\cos ^2 x}(-2 \cos x \sin x)-0 \\
&\therefore f^{\prime}(x)=0 \Rightarrow f(x)=\mathrm{C} \quad \text { (constant) }
\end{aligned}
$
Now, we find $f(x)$ at $x=\frac{\pi}{4}$
$
\begin{aligned}
\therefore \mathrm{I} &=\int_0^{1 / 2} \sin ^{-1} \sqrt{t} d t+\int_0^{1 / 2} \cos ^{-1} \sqrt{t} d t \\
&=\int_0^{1 / 2}\left(\sin ^{-1} \sqrt{t}+\cos ^{-1} \sqrt{t}\right) d t \\
&=\int_0^{1 / 2} \frac{\pi}{2} d t=\frac{\pi}{4}=\mathrm{C} \\
\therefore \quad f(x)=\frac{\pi}{4} \\
\therefore \text { Required integration }=\frac{\pi}{4}
\end{aligned}
$