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

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Question

(A) Let $I = \int \sin^{-1} x \cdot 1 \, dx$. Using integration by parts,$\int u \cdot v \, dx = u \int v \, dx - \int (u' \cdot \int v \, dx) \, dx$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \sin^{-1} x \cdot 1 \, dx$. Using integration by parts,$\int u \cdot v \, dx = u \int v \, dx - \int (u' \cdot \int v \, dx) \, dx$. Here,$u = \sin^{-1} x$ and $v = 1$. $I = \sin^{-1} x \cdot x - \int \frac{1}{\sqrt{1 - x^2}} \cdot x \, dx$. $I = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} \, dx$. To solve $\int \frac{x}{\sqrt{1 - x^2}} \, dx$,let $1 - x^2 = t$,then $-2x \, dx = dt$,so $x \, dx = -\frac{1}{2} dt$. $\int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \cdot \frac{t^{1/2}}{1/2} = -\sqrt{t} = -\sqrt{1 - x^2}$. Substituting this back into the expression for $I$: $I = x \sin^{-1} x - (-\sqrt{1 - x^2}) + c = x \sin^{-1} x + \sqrt{1 - x^2} + c$.

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

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