The value of int_0^{sin^2 x} sin^{-1} sqrt{t} | vedclass

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(C) Let $I = \int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \,dt + \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \,dt$. For the first integral,let $t = \sin^2 u$,then $d

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$\frac{\pi}{4}$

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(C) Let $I = \int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \,dt + \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \,dt$. For the first integral,let $t = \sin^2 u$,then $dt = 2 \sin u \cos u \,du = \sin 2u \,du$. When $t=0, u=0$ and when $t=\sin^2 x, u=x$. So,$\int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \,dt = \int_0^x u \sin 2u \,du$. For the second integral,let $t = \cos^2 v$,then $dt = -2 \cos v \sin v \,dv = -\sin 2v \,dv$. When $t=0, v=\pi/2$ and when $t=\cos^2 x, v=x$. So,$\int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \,dt = \int_{\pi/2}^x v (-\sin 2v) \,dv = \int_x^{\pi/2} v \sin 2v \,dv$. Thus,$I = \int_0^x u \sin 2u \,du + \int_x^{\pi/2} u \sin 2u \,du = \int_0^{\pi/2} u \sin 2u \,du$. Using integration by parts: $\int u \sin 2u \,du = u \left( -\frac{\cos 2u}{2} \right) - \int 1 \cdot \left( -\frac{\cos 2u}{2} \right) \,du = -\frac{u \cos 2u}{2} + \frac{\sin 2u}{4}$. Evaluating from $0$ to $\pi/2$: $\left[ -\frac{u \cos 2u}{2} + \frac{\sin 2u}{4} \right]_0^{\pi/2} = \left( -\frac{\pi/2 \cos \pi}{2} + 0 \right) - (0 + 0) = -\frac{\pi/2 (-1)}{2} = \frac{\pi}{4}$.

The value of $\int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \,dt + \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \,dt$ is

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