int frac{sin^{-1}x}{(1-x^2)^{3/2}} , dx = | vedclass

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Question

(A) Let $t = \sin^{-1}x$. Then $x = \sin t$ and $dx = \cos t \, dt$. Substituting these into the integral: $\int \frac{t}{(1-\sin^2 t)^{3/2}} \cos t \

Vedclass · Accepted Answer

$\frac{x}{\sqrt{1-x^2}} \sin^{-1}x + \frac{1}{2} \log(1-x^2) + c$

Vedclass · Answer

(A) Let $t = \sin^{-1}x$. Then $x = \sin t$ and $dx = \cos t \, dt$. Substituting these into the integral: $\int \frac{t}{(1-\sin^2 t)^{3/2}} \cos t \, dt = \int \frac{t}{(\cos^2 t)^{3/2}} \cos t \, dt = \int \frac{t}{\cos^3 t} \cos t \, dt = \int t \sec^2 t \, dt$. Using integration by parts: $\int t \sec^2 t \, dt = t 	an t - \int 	an t \, dt = t 	an t + \log|\cos t| + c$. Since $t = \sin^{-1}x$,then $	an t = \frac{x}{\sqrt{1-x^2}}$ and $\cos t = \sqrt{1-x^2}$. Substituting back: $= \sin^{-1}x \cdot \frac{x}{\sqrt{1-x^2}} + \log(\sqrt{1-x^2}) + c = \frac{x}{\sqrt{1-x^2}} \sin^{-1}x + \frac{1}{2} \log(1-x^2) + c$.

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

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