Question

Integrate the function: \(xsin^{-1}x\)

Answer 1

\(Let\) \(I\)=\(∫\)\(xsin^{-1}x\ dx\)

Taking as first function and x as second function and integrating by parts, we obtain

I= \(sin^{-1}x∫[x dx-∫{(\frac {d}{dx}sin^{-1}x)}∫x dx]dx\)

\(I= sin^{-1}x (\frac {x^2}{2})-∫\frac {1}{\sqrt {1-x^2}}.\frac {x^2}{2} dx\)

\(I= \frac {x^2sin^{-1}x}{2}+\frac 12∫\frac {-x^2}{\sqrt {1-x^2}} dx\)

I= \(\frac {x^2sin^{-1}x}{2}\) + \(\frac 12∫[{\frac {1-x^2}{\sqrt {1-x^2}}-\frac {1}{√1-x^2}}]dx\)

I= \(\frac {x^2sin^{-1}x}{2}\) + \(\frac 12∫[{\sqrt {1-x^2}-\frac {1}{\sqrt {1-x^2}}}]dx\)

I= \(\frac {x^2sin^{-1}x}{2}\) + \(\frac 12∫\frac x2{\sqrt {1-x^2\ }dx-∫\frac {1}{\sqrt {1-x^2}}}\ dx\)

I= \(\frac {x^2sin^{-1}x}{2}\) + \(\frac 12[{\frac x2\sqrt {1-x^2}+\frac 12sin^{-1}x-sin^{-1}x}]+C\)

I= \(\frac {x^2sin^{-1}x}{2}\) + \({\frac x4\sqrt {1-x^2}+\frac 14sin^{-1}x-\frac 12sin^{-1}x}+C\)

\(I= \frac 14(2x^2-1)sin^{-1}x+\frac x4\sqrt {1-x^2}+C\)