Integration C4

INT (2sin3xsin2x.dx)

${Integration by parts:}$

$\huge \int 2sin3xsin2x dx$

$\huge let u = 2sin3x$

$\huge let \frac{dv}{dx} = sin2x$

$\huge \Rightarrow \frac{du}{dx} = 6cos3x {and} v = -\frac{1}{2}cos2x$

$\huge \int 2sin3xsin2x dx = (2sin3x)(-\frac{1}{2}cos2x) - \int (6cos3x)(-\frac{1}{2}cos2x) dx$

$\huge -sin3xcos2x + \int 3(cos3x)(cos2x) dx$
It should be easy to solve from here.

Wouldn't it leave you with same problem as your left with two different trig functions of x and doing it by parts again doesn't seem its worth the mark. Does the identity 2sin(3x)sin(2x) --> Identity: 2sin(a)sin(b) = cos(a-b) - cos(a+b)
come under the spec for edexcel ??

This will work, it simply requires another integration by parts using the logical choice of $u$ and $v'$.
That said, I would still go with the trig identity.
Also try not to bump really old threads, if you have a question make a new thread.
@Zacken