What is the definite integral of (x^2-3x)cos2x from 0 to pi ?

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Answer 1 · 2018-05-04T14:36:16

The answer is #=pi/2#

First determine the definite integral by integration by parts #2times#

The integration by parts

#intuv'=uv-intu'v#

The integral is

#I=int(x^2-3x)cos2xdx#

#u=x^2-3x#, #=>#, #u'=2x-3#

#v'=cos2x#, #=>#, #v=1/2sin2x#

Therefore,

#I=1/2(x^2-3x)sin(2x)-int1/2(2x-3)sin2xdx#

Applying integration by parts a second time

#u=1/2(2x-3)#, #=>#, #u'=1#

#v'=sin2x#, #=>#, #v=-1/2cos(2x)#

#I=1/2(x^2-3x)sin(2x)-(-1/4(2x-3)cos(2x)+1/2intcos2xdx)#

#=1/2(x^2-3x)sin(2x)+1/4(2x-3)cos(2x)-1/2intcos2xdx#

#=1/2(x^2-3x)sin(2x)+1/4(2x-3)cos(2x)-1/4sin2x +C#

The definite integral is

#int_0^pi(x^2-3x)cos2xdx=[1/2(x^2-3x)sin(2x)+1/4(2x-3)cos(2x)-1/4sin2x]_0^pi#

#=(1/4(2pi-3))-(1/4*(-3))#

#=pi/2-3/4+3/4#

#=pi/2#