Question

Evaluate : 
`int xcos^-1x dx`

Answer 1

Let I = `int xcos^-1x  dx`
By using LIATE rule
= `int cos^-1 x  dx`
Integrating by parts
I = `cos^-1x ( x^2/2 ) - int[-1]/sqrt ( 1 - x^2 )(( x^2)/2) dx`

∴ I = `x^2/x cos^-1x + 1/2 int x^2/sqrt( 1 - x^2) dx`

= `x^2/2 cos^-1 x + 1/2 int [ 1 - (1- x^2 )]/[sqrt( 1 - x^2)]`

= `x^2/2 cos^-1x + 1/2 int [1]/[sqrt(1 - x^2)]dx - 1/2int sqrt( 1 - x^2) dx`

= `x^2/2 cos^-1x + 1/2 sin^-1x - 1/2[ x/2 sqrt( 1 - x^2) + 1/2 sin^-1 x ]`

= `x^2/2 cos^-1x - x/4 sqrt( 1 - x^2 ) + 1/4 sin^-1x + C`