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

1 Answer
maganbhai P.
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#

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