Find the exact value of int_0^(1/2pi)x^2sin2x dx?

1 Answer
Jul 17, 2018

int_0^(pi/2) x^2*sin2xdx=pi^2/8-1/2

Explanation:

Here,

I=int_0^(pi/2) x^2*sin2xdx

Using Integration by parts:

I=[x^2*(-cos(2x)/2)]_0^(pi/2) -int_0^(pi/2)2x(-cos(2x)/2)dx

=[pi^2/4(-cospi/2)-0]+int_0^(pi/2)xcos2xdx

Again using Integration by parts:

=[pi^2/4(-(-1)/2)]+[x(sin(2x)/2)]_0^(pi/2) - int_0^(pi/2)1(sin(2x)/2)dx

=pi^2/8+[pi/2*sinpi/2-0]-1/2[-cos(2x)/2]_0^(pi/2)

=pi^2/8+0+1/4[cospi-cos0]

=pi^2/8+1/4[-1-1]

:.I=pi^2/8-1/2