Find the integral of following?

Question

#int_0^(pi/2) xcos(x)#

879 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-21T12:58:27

#I=pi/2-1#

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#