Find the integral of following?

int_0^(pi/2) xcos(x)

1 Answer
Apr 21, 2018

I=pi/2-1

Explanation:

Here,

I=int_0^(pi/2) xcosxdx

"Using "color(blue)"Integration by Parts"

intu*vdx=uintvdx-int(u'intvdx)dx

Let, u=x and v=cosx=>u'=1 and intvdx=sinx

So,

I=[x*sinx]_0^(pi/2)-int_0^(pi/2)(1*sinx)dx

=[pi/2sin(pi/2)-0]-int_0^(pi/2)sinxdx

=[pi/2(1)]-[-cosx]_0^(pi/2)

=pi/2+[cos(pi/2)-cos0]

=pi/2+0-1

=pi/2-1