#\int x^2sin^{-1}xdx#
Applying integration by parts:
=#int \uv'=uv-\int u'v# ...(i)
=#u=\sin^-1 (x),u'=\frac{1}{\sqrt{1-x^2}},v'=x^2,v=\frac{x^3}{3}#
=#u=\sin^-1(x)# #v'=x^2#
=#u'=\frac{d}{dx}(\sin^-1 (x))# = #\frac{1}{\sqrt{1-x^2}}#
(Applying common derivative #\frac{d}{dx}(\sin^-1(x))=\frac{1}{\sqrt{1-x^2}}#
=#v=\int x^2dx# = #x^3/3#
=#u=\sin^-1 (x),u'=\frac{1}{\sqrt{1-x^2}},v'=x^2,v=\frac{x^3}{3}#
=#sin^-1 (x)\frac{x^3}{3}-\int \frac{1}{\sqrt{1-x^2}}\frac{x^3}{3}dx# (from i)
=#\frac{1}{3}x^3\sin^-1 (x)-\int \frac{x^3}{3\sqrt{1-x^2}}dx#
We have # \int \frac{x^3}{3\sqrt{1-x^2}}dx# = #\frac{1}{3}(\frac{1}{3}(1-x^2)^{\frac{3}{2}}-\sqrt{1-x^2})#
(Taking the constant out #\int a\cdot f(x)dx=a\cdot \int f(x)dx# and applying sum rule #\int f(x)\pm g(x\)dx=\int f(x)dx\pm \int g(x)dx#)
=#\frac{1}{3}x^3\sin^-1 (x)# - #\frac{1}{3}(\frac{1}{3}(1-x^2)^{\frac{3}{2}}-\sqrt{1-x^2})#
Simplifying it
=#\frac{1}{3}x^3sin^-1 (x)+\frac{1}{3}(\sqrt{1-x^2}-\frac{1}{3}(1-x^2)^{\frac{3}{2}})#
=#\frac{1}{3}x^3\sin^-1 (x)+\frac{1}{3}(\sqrt{1-x^2}-\frac{1}{3}(1-x^2)^{\frac{3}{2}})+C#
