Question

CLass 12th


$\implies \pink{\int \: x \sin {}^{ - 1} x }$


Don't spam!!​

Answer 1

Given:

int xsin-¹ x

$\displaystyle \: M = \int x\sin^{-1}(x)dx$

Solution:

$\displaystyle \: M = \int x\sin^{-1}(x)dx$

Integrating by parts

${{\displaystyle: \implies \sin^{-1}(x)\int x \cdot dx}}$

${{\displaystyle : \implies \int\Bigg[\frac{d}{dx}{\sin^{-1} {x}}\int x dx\Bigg]dx}}$

${\displaystyle:\implies\frac{x^2\sin^{-1}(x)}{2}\frac{1}{2}\int\Bigg(\frac{x^2}{\sqrt{1 x^2}}\Bigg)dx \: \: \: \: \: \: - (1)}$

Now we have to solve

$:\implies \: l=\displaystyle\int\Bigg(\frac{x^2}{\sqrt{1x^2}}\Bigg)dx$

Substitution

$\displaystyle \: 1-x^2=t^2$

$: \implies \displaystyle \: -x\cdot dx-t\cdot dt$

$:\implies \displaystyle \: dx=-\frac{t\cdot dt}{x}$

$:\implies l=\displaystyle -\int\frac{x^2} {t}\cdot\frac{t\cdot dt}{x}$

$: \implies -\displaystyle x\cdot dt$

$: \implies \displaystyle \sqrt{1-t^2}\cdot dt$

${\displaystyle {: \implies-\Bigg[\frac{t}{2}\sqrt{1 t^2} + \frac{1}{2}\sin^{-1}{(t)}\Bigg]}}$

Substituting

$\displaystyle \: let^2=1-x^2$

${\displaystyle {:\implies-\Bigg[\frac{\sqrt{1-x^2}} {2}x+\frac{1}{2}\sin^{-1}{(\sqrt{1-x^2}}}\Bigg]}$

Putting I in equation (1)

$:\implies\displaystyle \int x\sin^{-1}(x)dx=$

${ \displaystyle{:\implies \frac{1}{2} \Bigg[\frac{\sqrt{1-x^2}}{2}x+}\frac{1}{2}\sin^{-1}{(\sqrt{1-x^2})}\Bigg] + }{\displaystyle\frac{x^2\sin^{-1} (x)}{2}+}constant$

Answer 2

$\huge\bold{\color{aqua}{Solution :-}}$

$\quad\bold{\int \: x \sin {}^{ - 1} x }$

$\bold{=sin^{-1}x.\frac{x²}{2}-\int \:\frac{1}{\sqrt{1-x²}}.\frac{x²}{2}\:dx}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{1}{2}\int \:\frac{-x²}{\sqrt{1-x²}}\:dx}$

$\bold{\purple{\text{On adding and subtracting 1}}}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{1}{2}\int \:\frac{-x²+1-1}{\sqrt{1-x²}}\:dx}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{1}{2}\int \:\frac{1-x²}{\sqrt{1-x²}}dx\:-\frac{1}{2}\int \:\frac{1}{\sqrt{1-x²}}dx}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{1}{2}\int \:\frac{\red{\cancel{\blue{1-x²}}}^{\green{\:\sqrt{1-x²}}}}{\red{\cancel{\blue{\sqrt{1-x²}}}}}dx\:-\frac{1}{2}sin^{-1}x}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{1}{2}\int\sqrt{1-x²}dx\:-\frac{1}{2}sin^{-1}x}$

$\small\bold{=\frac{x²}{2}.sin^{-1}x+\frac{1}{2}\bigg(\frac{x}{2}\sqrt{1-x²}+\frac{1}{2}sin^{-1}x\bigg)-\frac{1}{2}sin^{-1}x+C}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{x}{4}\sqrt{1-x²}+\frac{1}{4}sin^{-1}x-\frac{1}{2}sin^{-1}x+C}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{x}{4}\sqrt{1-x²}+\bigg(\frac{1}{4}-\frac{1}{2}\bigg)sin^{-1}x+C}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{x}{4}\sqrt{1-x²}+\frac{1}{4}sin^{-1}x-\frac{1}{2}sin^{-1}x+C}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{x}{4}\sqrt{1-x²}+\bigg(\frac{1-2}{4}\bigg)sin^{-1}x+C}$

$\bold{=\frac{x²}{2}.sin^{-1}x+\frac{x}{4}\sqrt{1-x²}-\frac{1}{4}sin^{-1}x+C}$

$\bold{=\frac{x²}{2}.sin^{-1}x-\frac{1}{4}sin^{-1}x+\frac{x}{4}\sqrt{1-x²}+C}$

$\bold{=\bigg(\frac{x²}{2}-\frac{1}{4}\bigg)sin^{-1}x+\frac{x}{4}\sqrt{1-x²}+C}$

$\bold{=\green{\boxed{\pink{\bigg(\frac{2x²-1}{4}\bigg)sin^{-1}x+\frac{x}{4}\sqrt{1-x²}+C}}}}$