3. int_0^{frac{1}{2}} frac{x sin^{-1} x}{sqrt | vedclass

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(B) Let $I = \int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1-x^2}} dx$. Substitute $\sin^{-1} x = t$,then $x = \sin t$ and $dx = \cos t dt$. When $x = 0$,$t

Vedclass · Accepted Answer

$\left(\frac{1}{2} - \frac{\sqrt{3}}{12} \pi\right)$

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(B) Let $I = \int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1-x^2}} dx$. Substitute $\sin^{-1} x = t$,then $x = \sin t$ and $dx = \cos t dt$. When $x = 0$,$t = 0$. When $x = 1/2$,$t = \pi/6$. Substituting these into the integral: $I = \int_0^{\pi/6} \frac{\sin t \cdot t}{\sqrt{1-\sin^2 t}} \cdot \cos t dt = \int_0^{\pi/6} t \sin t dt$. Using integration by parts: $\int u dv = uv - \int v du$. Let $u = t$ and $dv = \sin t dt$,then $du = dt$ and $v = -\cos t$. $I = [t(-\cos t)]_0^{\pi/6} - \int_0^{\pi/6} (-\cos t) dt$ $I = [-t \cos t + \sin t]_0^{\pi/6}$ $I = [-\frac{\pi}{6} \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6})] - [0 + \sin(0)]$ $I = [-\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}] = \frac{1}{2} - \frac{\sqrt{3}}{12} \pi$.

$3. \int_0^{\frac{1}{2}} \frac{x \sin^{-1} x}{\sqrt{1-x^2}} dx =$

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