🎥 Video solution / Text Solution of this question is given below:
Use the identity:
$\sin^{-1}y+\cos^{-1}y=\dfrac{\pi}{2}$
So:
$\cos^{-1}\sqrt{t}=\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}$
Now substitute:
$I=\displaystyle\int_0^{\sin^2 x}\sin^{-1}\sqrt{t} dt + \int_0^{\cos^2 x} \left(\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}\right) dt$
$I=\dfrac{\pi}{2}\cos^2 x + \int_0^{\sin^2 x}\sin^{-1}\sqrt{t}dt - \int_0^{\cos^2 x}\sin^{-1}\sqrt{t} dt$
Combine integrals:
$I=\dfrac{\pi}{2}\cos^2 x + \int_{\cos^2 x}^{\sin^2 x}\sin^{-1}\sqrt{t} dt$
But:
$\sin^2 x + \cos^2 x = 1$
Limits become from $1$ to $0$:
$I=\dfrac{\pi}{2}(1 - \sin^2 x) + \int_{1}^{0}\sin^{-1}\sqrt{t} dt$
$I=\dfrac{\pi}{2} - \int_0^{1}\sin^{-1}\sqrt{t} dt$
Let $u=\sqrt{t}$,
$dt=2udu$:
$\displaystyle \int_0^{1}\sin^{-1}\sqrt{t} dt = 2\int_0^{1} u\sin^{-1}udu$
Standard result:
$\displaystyle 2\int_0^{1} u\sin^{-1}u du = \dfrac{\pi}{4}$
Thus:
$I = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$