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

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

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$\left( {\frac{{{x^2}}}{2} - \frac{1}{4}} \right){\sin ^{ - 1}}x + \frac{x}{4}\sqrt {1 - {x^2}}  + c$

Vedclass · Answer

(A) Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = \sin^{-1}x$ and $v = x$. $\int x \sin^{-1}x dx = \frac{x^2}{2} \sin^{-1}x - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} dx + c$ $= \frac{x^2}{2} \sin^{-1}x - \frac{1}{2} \int \frac{-(1-x^2)+1}{\sqrt{1-x^2}} dx + c$ $= \frac{x^2}{2} \sin^{-1}x + \frac{1}{2} \int \sqrt{1-x^2} dx - \frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} dx + c$ Using the standard integral $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$,we get: $= \frac{x^2}{2} \sin^{-1}x + \frac{1}{2} \left( \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}x \right) - \frac{1}{2}\sin^{-1}x + c$ $= \frac{x^2}{2} \sin^{-1}x + \frac{x}{4}\sqrt{1-x^2} + \frac{1}{4}\sin^{-1}x - \frac{1}{2}\sin^{-1}x + c$ $= \left( \frac{x^2}{2} - \frac{1}{4} \right) \sin^{-1}x + \frac{x}{4}\sqrt{1-x^2} + c$.

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

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