`inte^2thetasin(3theta)d theta`

If f(x) and g(x) are differentiable functions, then

`intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx`

If we rewrite f(x)=u and g'(x)=v, then

`intuvdx=uintvdx-int(u'intvdx)dx`

Let's integrate using the above method of integration by parts,

Let `u=e^2theta, v=sin(3theta)`

`inte^(2theta)sin(3theta)d theta=e^(2theta)intsin(3theta)d theta-int(d/(d theta)(e^(2theta))intsin(3theta)d theta)d theta`

`=e^(2theta)(-1/3cos(3theta)-int(e^(2theta)2(-1/3cos(3theta))d theta`

`=-1/3e^(2theta)cos(3theta)+2/3inte^(2theta)cos(3theta)d theta`

apply again integration by parts,

`=-1/3e^(2theta)cos(3theta)+2/3(e^(2theta)*intcos(3theta)d theta-int(e^(2theta)*2intcos(3theta) d theta)`

`=-1/3e^(2theta)cos(3theta)+2/3(e^(2theta)1/3sin(3theta)-int2e^(2theta)(sin(3theta)/3)d theta)`

`=-1/3e^(2theta)cos(3theta)+2/9e^(2theta)sin(3theta)-4/9inte^(2theta)sin(3theta)d theta`

Isolate `inte^(2theta)sin(3theta)d theta`

`(1+4/9)inte^(2theta)sin(3theta)d theta=-1/3e^(2theta)cos(3theta)+2/9e^(2theta)sin(3theta)`

`inte^(2theta)sin(3theta)d theta=9/13(-1/3e^(2theta)cos(3theta)+2/9e^(2theta)sin(3theta))`

`=-3/13e^(2theta)cos(3theta)+2/13e^(2theta)sin(3theta)`

add a constant C to the solution,

`=-3/13e^(2theta)cos(3theta)+2/13e^(2theta)sin(3theta)+C`

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