Question

Integrate the function.
$\int xsi{n}^{-1}xdx$

Answer 1

Let $I=\int xsi{n}^{-1}xdx$
On taking $si{n}^{-1}x$ as first function and x as second function and integrating by parts, we get
$(\because$ Inverse functions comes before, algebraic functions in ILATE)
$I=si{n}^{-1}x\int xdx-\int \left[\frac{d}{dx}\right(si{n}^{-1}x)\int xdx]dx=si{n}^{-1}x.\frac{x^2}{2}-\int [\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2}]dx=\frac{x^2}{2}.si{n}^{-1}x+\int [\frac{1-1-x^2}{\sqrt{1-x^2}}.\frac{1}{2}]dx$
(Add and subtract 1 in numerator of second term)
$=\frac{x^2}{2}.si{n}^{-1}x-\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}dx+\frac{1}{2}\int \frac{1-x^2}{\sqrt{1-x^2}}dx=\frac{x^2}{2}.si{n}^{-1}x-\frac{1}{2}si{n}^{-1}x+\frac{1}{2}\int \sqrt{1-x^2}dx\Rightarrow I=\frac{x^2}{2}.si{n}^{-1}x-\frac{1}{2}si{n}^{-1}x+\frac{1}{2}[\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}si{n}^{{}_{-}1}x]+C\Rightarrow I=si{n}^{-1}x[\frac{x^2}{2}-\frac{1}{4}]+\frac{1}{4}x\sqrt{1-x^2}+C\Rightarrow I=\frac{si{n}^{-1}}{4}(2x^2-1)+\frac{1}{4}x\sqrt{1-x^2}+C$