int_0^1 x operatorname{Sin}^{-1} x , dx = | vedclass

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(A) To evaluate the integral $I = \int_0^1 x \operatorname{Sin}^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u

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$\frac{\pi}{8}$

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(A) To evaluate the integral $I = \int_0^1 x \operatorname{Sin}^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \operatorname{Sin}^{-1} x$ and $dv = x \, dx$. Then $du = \frac{1}{\sqrt{1-x^2}} \, dx$ and $v = \frac{x^2}{2}$. Applying the formula: $I = \left[ \frac{x^2}{2} \operatorname{Sin}^{-1} x ight]_0^1 - \int_0^1 \frac{x^2}{2 \sqrt{1-x^2}} \, dx$. Evaluating the first part: $\left[ \frac{1^2}{2} \operatorname{Sin}^{-1}(1) - 0 ight] = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$. For the second part,let $J = \int_0^1 \frac{x^2}{2 \sqrt{1-x^2}} \, dx$. Substitute $x = \sin 	heta$,$dx = \cos 	heta \, d	heta$. $J = \frac{1}{2} \int_0^{\pi/2} \frac{\sin^2 	heta \cos 	heta}{\cos 	heta} \, d	heta = \frac{1}{2} \int_0^{\pi/2} \sin^2 	heta \, d	heta$. Using $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$: $J = \frac{1}{4} \int_0^{\pi/2} (1 - \cos 2	heta) \, d	heta = \frac{1}{4} \left[ 	heta - \frac{\sin 2	heta}{2} ight]_0^{\pi/2} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8}$. Thus,$I = \frac{\pi}{4} - \frac{\pi}{8} = \frac{\pi}{8}$.

$\int_0^1 x \operatorname{Sin}^{-1} x \, dx =$

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