int frac{sin ^{-1} sqrt{x}-cos ^{-1} sqrt{x}} | vedclass

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(B) Let $I = \int \frac{\sin ^{-1} \sqrt{x} - \cos ^{-1} \sqrt{x}}{\sqrt{x}(\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x})} dx$. Since $\sin ^{-1} \sqrt{x

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$\frac{8}{\pi}\left(\sqrt{x} \sin ^{-1} \sqrt{x}+\sqrt{1-x}\right)-2 \sqrt{x}+C$

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(B) Let $I = \int \frac{\sin ^{-1} \sqrt{x} - \cos ^{-1} \sqrt{x}}{\sqrt{x}(\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x})} dx$. Since $\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x} = \frac{\pi}{2}$,we have $\cos ^{-1} \sqrt{x} = \frac{\pi}{2} - \sin ^{-1} \sqrt{x}$. Substituting this,$I = \int \frac{\sin ^{-1} \sqrt{x} - (\frac{\pi}{2} - \sin ^{-1} \sqrt{x})}{\sqrt{x}(\frac{\pi}{2})} dx = \frac{2}{\pi} \int \frac{2 \sin ^{-1} \sqrt{x} - \frac{\pi}{2}}{\sqrt{x}} dx$. Let $\sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,so $\frac{dx}{\sqrt{x}} = 2 dt$. $I = \frac{2}{\pi} \int (2 \sin ^{-1} t - \frac{\pi}{2}) (2 dt) = \frac{8}{\pi} \int \sin ^{-1} t dt - 2 \int dt$. Using $\int \sin ^{-1} t dt = t \sin ^{-1} t + \sqrt{1-t^2} + C$,we get: $I = \frac{8}{\pi} (t \sin ^{-1} t + \sqrt{1-t^2}) - 2t + C$. Substituting $t = \sqrt{x}$: $I = \frac{8}{\pi} (\sqrt{x} \sin ^{-1} \sqrt{x} + \sqrt{1-x}) - 2\sqrt{x} + C$.

$\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sqrt{x}\left(\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}\right)} d x=$

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