Show $\int_0^{2\pi}\cos(n\phi')\cos^l(\phi-\phi')\mathrm{d}\phi=\frac{2\pi}{2^l}\cos(l\phi)\delta_{l,n}$ I have to show $$\int_0^{2\pi}\cos(n\phi')\cos^l(\phi-\phi')\mathrm{d}\phi=\frac{2\pi}{2^l}\cos(l\phi)\delta_{l,n}$$
where $l,n$ are positive integers such that $l\leq n$
I'm supposed to use the fact 
$$ \int_0^{2\pi}e^{i(l-n)\phi}\mathrm{d}\phi=2\pi\delta_{l,n}$$
But I'm really lost. I tried to rewrite the cosines as the real part of $e^{in\phi}$ et cetera and trying to expand the power with the binomial theorem but it didn't work. I don't see any other way to obtain a useful way to use that identity. Any input will be appreciated

Use $\cos x=(e^{ix}+e^{-ix})/2$ to rewrite
$$
\int_0^{2\pi}\cos(n\phi')\cos^l(\phi-\phi')\mathrm{d}\phi'=\frac{1}{2^{l+1}}\int_0^{2\pi}(e^{in\phi'}+e^{-in\phi'})(e^{i(\phi-\phi')}+e^{-i(\phi-\phi')})^l d\phi'\ .
$$
Then use the binomial theorem to rewrite
$$
\frac{1}{2^{l+1}}\sum_{k=0}^l {l\choose k}\int_0^{2\pi}(e^{in\phi'}+e^{-in\phi'})e^{ik(\phi-\phi')}e^{-i(l-k)(\phi-\phi')} d\phi'=\frac{1}{2^{l+1}}\sum_{k=0}^l {l\choose k} e^{i\phi(2k-l)}2\pi(\delta_{n-k+(l-k),0}+\delta_{-n-k+(l-k),0})=\frac{2\pi}{2^{l+1}}\left[{l\choose (l+n)/2}e^{in\phi}+{l\choose (l-n)/2}e^{-in\phi}\right]\ ,
$$
from which you should be able to complete.
