Question

need help evaluating.. ∫sin(3x)cos(x)dx

Answer 1

I know I have to split it up. But having problems doing so. I am not sure how to split up sin(3x), any ideas? I am sure this is an easy one I am just new

Answer 2

no ideas?

Answer 3

maybe you should start by making your odd number power even like this:

\[\int\limits_{?}^{?}\sin^2(x)\cos^2(x)\sin(x) dx\]

Answer 4

and then use trig substitution \[\sin^2x=1-\cos^2x\]

Answer 5

\[\sin(3x) = \sin(2x +x) = \sin(2x)\cos(x)+\sin(x)\cos(2x)\]
\[\sin(2x) = 2\sin(x)\cos(x)\]
\[\cos(2x) = 2\cos^{2}x -1\]
\[\rightarrow \sin(3x) = 4\sin(x)\cos^{2}x - \sin(x)\]
then use substitution
u = cos x
du = -sin x dx

Answer 6

an easier way is to remember this all-too-forgotten trig identity:\[\sin\alpha\cos\beta=\frac12[(\sin(\alpha+\beta)+\sin(\alpha-\beta)]\]I don't know why they don't teach this very often...