Question

Integrate the function : $x\ sin^{-1} x$

Answer 1

Please see the attachment

Answer 2

This integration question is based on integration by part,

if f(x) is first function and g(x) is second function then,

$\int{f(x).g(x)}\,dx=f(x)\int{g(x)}\,dx-\int{f'(x)\int{g(x)}\,dx}\,dx$

here in question,

first function, f(x) = sin-¹x

and second function, g(x) = x

so, $\int{xsin^{-1}x}\,dx=sin^{-1}\int{x}\,dx-\int{\frac{d(sin^{-1}x)}{dx}\int{x}\,dx}\,dx$

= $sin^{-1}x\frac{x^2}{2}-\int{\frac{1}{\sqrt{1-x^2}}\frac{x^2}{2}}\,dx$

= $sin^{-1}x\frac{x^2}{2}-\frac{1}{2}\int{\frac{x^2}{\sqrt{1-x^2}}}\,dx$

= $sin^{-1}x\frac{x^2}{2}+\frac{1}{2}\int{\frac{(1-x^2-1}{\sqrt{1-x^2}}}\,dx$

= $sin^{-1}x\frac{x^2}{2}+\frac{1}{2}\int{\sqrt{1-x^2}}\,dx-\frac{1}{2}\int{\frac{1}{\sqrt{1-x^2}}}\,dx$

= $sin^{-1}x\frac{x^2}{2}+\frac{1}{2}\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}sin^{-1}x\right]-\frac{1}{2}sin^{-1}x+C$

= $sin^{-1}x\frac{x^2}{2}+\frac{x}{4}\sqrt{1-x^2}-\frac{1}{4}sin^{-1}x+C$