Question

Please help me integrate the following problem:

sin(4x)sin(8x)dx

Answer 1

\[\int\limits_{}^{}\sin(4x)\sin(8x)dx\]

Answer 2

Ok so let's recall a couple of identities that include something like what you have inside that integral
The cos sum and difference identity has something that looks like that.
\[\cos(4x+8x)=\cos(4x)\cos(8x)-\sin(4x)\sin(8x)\]
\[\cos(4x-8x)=\cos(4x)\cos(8x)+\sin(4x)\sin(8x)\]
now we can solve this system of sin(4x)sin(8x) but multiplying the first identity by negative 1 and then adding the equals together like so:
\[-\cos(4x+8x)=-\cos(4x)\cos(8x)+\sin(4x)\sin(8x)\]
\[\cos(4x-8x)=\cos(4x)\cos(8x)+\sin(4x)\sin(8x)\]
now add together
\[-\cos(4x+8x)+\cos(4x-8x)=2\sin(4x)\sin(8x)\]
or just
\[\sin(4x)\sin(8x)=\frac{1}{2}(-\cos(12x)+\cos(-4x))\]

Answer 3

There is actually an identity for this but I'm just showing you the identity using other identities. :)