int_0^{1/sqrt{2}} frac{sin^{-1}x}{(1-x^2)^{3/ | vedclass

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Question

(B) Let $I = \int_0^{1/\sqrt{2}} \frac{\sin^{-1}x}{(1-x^2)^{3/2}} dx$. Substitute $\sin^{-1}x = t$,which implies $x = \sin t$ and $dx = \cos t \, dt$.

Vedclass · Accepted Answer

$\frac{\pi}{4} - \frac{1}{2}\log 2$

Vedclass · Answer

(B) Let $I = \int_0^{1/\sqrt{2}} \frac{\sin^{-1}x}{(1-x^2)^{3/2}} dx$. Substitute $\sin^{-1}x = t$,which implies $x = \sin t$ and $dx = \cos t \, dt$. When $x = 0$,$t = 0$. When $x = 1/\sqrt{2}$,$t = \pi/4$. The integral becomes $I = \int_0^{\pi/4} \frac{t \cos t}{(1-\sin^2 t)^{3/2}} dt = \int_0^{\pi/4} \frac{t \cos t}{\cos^3 t} dt = \int_0^{\pi/4} t \sec^2 t \, dt$. Using integration by parts: $\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = \sec^2 t \, dt$. Then $du = dt$ and $v = 	an t$. $I = [t 	an t]_0^{\pi/4} - \int_0^{\pi/4} 	an t \, dt$. $I = [\frac{\pi}{4} 	an(\frac{\pi}{4}) - 0] - [\log|\sec t|]_0^{\pi/4}$. $I = \frac{\pi}{4}(1) - [\log(\sec \frac{\pi}{4}) - \log(\sec 0)]$. $I = \frac{\pi}{4} - [\log(\sqrt{2}) - \log(1)] = \frac{\pi}{4} - \log(2^{1/2}) = \frac{\pi}{4} - \frac{1}{2} \log 2$.

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

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