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Q.
The value of $ \int\limits _0^{\sin^2x} \sin ^{-1} \sqrt {t} \, dt + \int\limits _0 ^{\cos^2x} \cos^{-1} \sqrt {t} \, {dt}$ is
Integrals
Solution:
Put $t = sin^{2} y$ in the integral
$\int\limits_{0}^{sin^2 \,x} sin^{-1} \sqrt{t}\,dt $
$\therefore dt=2\, sin\, y\, dy=sin\, 2y\, dy$
Put $t =cos^{2} u$ in the integral $\int\limits_{0}^{cos^2 \,x} cos^{-1} \sqrt{t}\,dt$
$\therefore dt =-2$ cos\, u sin\, u du = - sin\, 2u \,du
$\therefore $ the given result $=\int\limits_{0}^{x} y sin 2y dy-\int\limits_{\pi /2}^{x}u sin 2u du$
$=\int\limits_{0}^{\pi/ 2} y sin 2y dy+\int\limits_{\pi /2}^{x}y sin 2y dy-\int\limits_{\pi /2}^{x}u sin 2u du$
$=\int\limits^{\pi /2}_{0} y sin 2y dy$
$=\left|y\left(\frac{-cos\,2y}{2}\right)\right|_{0}^{\pi/ 2}-\int\limits_{0}^{\pi/ 2}1: -\frac{cos\,2y}{2}dy$
$=-\frac{\pi}{4}\left(-1\right)+\frac{1}{2}\left|\frac{sin\,2y}{2}\right|_{0}^{\pi /2}$
$=\frac{\pi}{4}$