Question

$\underset{0}{\overset{1}{\int }}{x}^2{\sin }^{-1}xdx$

Answer 1

Consider the given integral.

$I={\int }_0^1{x}^2{\sin }^{-1}xdx$

We know that

$\int uvdx=u\int vdx-\int (\frac{d}{dx}\left(u\right)\int vdx)dx$

Therefore,

$I={\left[{\sin }^{-1}x\left(\frac{x^3}{3}\right)\right]}_0^1-{\int }_0^1(\frac{1}{\sqrt{1-x^2}}\times \left(\frac{x^3}{2}\right))dx$

$I=[{\sin }^{-1}\left(1\right)\left(\frac{1^3}{3}\right)-{\sin }^{-1}\left(0\right)\left(\frac{0^3}{3}\right)]-\frac{1}{2}{\int }_0^1\left(\frac{x^3}{\sqrt{1-x^2}}\right)dx$

$I=[\frac{1}{3}{\sin }^{-1}\left(\sin \frac{\pi }{2}\right)-0]-\frac{1}{2}{\int }_0^1\left(\frac{x^3}{\sqrt{1-x^2}}\right)dx$

$I=\frac{\pi }{6}-\frac{1}{2}{\int }_0^1\left(\frac{x^3}{\sqrt{1-x^2}}\right)dx$

Let $t=1-x^2$

$\frac{dt}{dx}=0-2x$

$-\frac{dt}{2}=xdx$

Therefore,

$I=\frac{\pi }{6}-\frac{1}{2}{\int }_1^0\left(\frac{(1-t)}{\sqrt{t}}\right)(-\frac{dt}{2})$

$I=\frac{\pi }{6}+\frac{1}{4}{\int }_1^0\left(\frac{(1-t)}{\sqrt{t}}\right)dt$

$I=\frac{\pi }{6}+\frac{1}{4}{\int }_1^0\frac{1}{\sqrt{t}}dt-\frac{1}{4}{\int }_1^0\sqrt{t}dt$

$I=\frac{\pi }{6}+\frac{1}{4}{\left(2\sqrt{t}\right)}_1^0-\frac{1}{4}{\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)}_1^0$

$I=\frac{\pi }{6}+\frac{1}{4}(2\sqrt{0}-2\sqrt{1})-\frac{1}{4}\left(\frac{2}{3}(0^{\frac{3}{2}}-1^{\frac{3}{2}})\right)$

$I=\frac{\pi }{6}+\frac{1}{4}(-2)-\frac{1}{4}(-\frac{2}{3})$

$I=\frac{\pi }{6}-\frac{1}{2}+\frac{1}{6}$

$I=\frac{\pi -3+1}{6}$

$I=\frac{\pi -2}{6}$

Hence, this is the answer.