int frac{x sin^{-1} x}{sqrt{1 - x^2}} , dx = | vedclass

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Question

(A) Let $I = \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx$. Substitute $t = \sin^{-1} x$,then $x = \sin t$ and $dt = \frac{1}{\sqrt{1 - x^2}} \, dx

Vedclass · Accepted Answer

$x - \sqrt{1 - x^2} \sin^{-1} x + c$

Vedclass · Answer

(A) Let $I = \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx$. Substitute $t = \sin^{-1} x$,then $x = \sin t$ and $dt = \frac{1}{\sqrt{1 - x^2}} \, dx$. The integral becomes $I = \int t \sin t \, dt$. Using integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = t$ and $dv = \sin t \, dt$. Then $du = dt$ and $v = -\cos t$. $I = -t \cos t - \int (-\cos t) \, dt = -t \cos t + \sin t + c$. Since $t = \sin^{-1} x$,we have $\sin t = x$ and $\cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2}$. Substituting these back,we get $I = -(\sin^{-1} x) \sqrt{1 - x^2} + x + c = x - \sqrt{1 - x^2} \sin^{-1} x + c$.

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

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