Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 17

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