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

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(B) Let $t = \sin^{-1} x$. Then $dt = \frac{1}{\sqrt{1 - x^2}} dx$. When $x = 0$,$t = 0$. When $x = 1/2$,$t = \sin^{-1}(1/2) = \pi/6$. Also,$x = \sin

Vedclass · Accepted Answer

$\frac{1}{2} - \frac{\sqrt{3} \pi}{12}$

Vedclass · Answer

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

$\int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx = $

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