Find $\displaystyle\int_0^\pi \cos^4\theta \sin^3\theta~d\theta$ using de Moivre's theorem.
So I need to find and expression for $\cos^4\theta \sin^3\theta$ in terms of multiple angles. I know that $2\cos\theta = z + z^{-1}$ and $2i\sin\theta = z-z^{-1}$ and my original thought was to work out $\cos^4\theta$ and $\sin^3\theta$ and then multiply my two expressions but I've now realised that this is something I won't be able to easily integrate. In addition to this the expression for $\cos^4\theta \sin^3\theta$ can be expression entirely as multiple angles of $\sin$. This has left me unsure on working out a expression for $\cos^4\theta \sin^3\theta$.
Any help would be appreciated
Best Answer
Indeed, we have $$ \cos^4\theta\sin^3\theta=\frac{3\sin\theta + 3 \sin 3\theta - \sin 5\theta - \sin 7\theta}{64} $$ which you can deduce from de Moivre's $(\cos x+i\sin x)^n=\cos nx+i\sin nx$ and Euler's $e^{ix}=\cos\theta+i\sin x$ or just by repeated product-to-sum. Now integrate from 0 to $\pi$.