Question #77e15

1 Answer
Feb 21, 2016

\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

Explanation:

\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