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

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Answer 1 · 2018-07-17T18:59:17

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

Here,

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

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

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

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

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

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

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

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

Find the exact value of int_0^(1/2pi)x^2sin2x dxint_0^(1/2pi)x^2sin2x dx?