Integral of $\big((1+\cos(x))\sin(x)\big)^2$

Question

What is $$\int \big((1+\cos(x))\sin(x)\big)^2dx$$ ?

Answer 1

use $$\begin{align}(1 + \cos t)^2\sin^2 t &= \sin^2 t + 2\sin^2 t \cos t + \sin^2 t \cos^2 t\\ &=\frac 12 - \frac 12 \cos 2t + 2\sin^2 t \cos t + \frac 18 - \frac 18 \cos 4t \\ &=\frac 58 - \frac 12 \cos 2t + 2\sin^2 t \cos t - \frac 18 \cos 4t \end{align}$$

now we can integrate $$\int (1 + \cos t)^2\sin^2 t\, dt = \frac 58 t - \frac 14 \sin 2t + \frac 23 \sin^3 t - \frac 1{32} \sin 4t + C $$

Answer 2

Use $\cos x=(e^{ix}+e^{-ix})/2$, $\sin x=-i(e^{ix}-e^{-ix})/2$

You get $$ -\frac{1}{16}\int\bigl((2+e^{ix}+e^{-ix})(e^{ix}-e^{-ix})\bigr)^2\,dx $$ Let's simplify the internal product: $$ (2+e^{ix}+e^{-ix})(e^{ix}-e^{-ix})= 2e^{ix}-2e^{-ix}+e^{2ix}-1+1-e^{-2ix} $$ and squaring it gives $$ 4e^{2ix}+4e^{-2ix}+e^{4ix}+e^{-4ix}-8+4e^{3ix}-4e^{-ix} -4e^{ix}+4e^{-3ix}-2 $$ that we can reorder as $$ e^{4ix}+e^{-4ix}+4e^{3ix}+4e^{-3ix}+4e^{2ix}+4e^{-2ix}-4e^{ix}-4e^{-ix}-10 $$ or, switching back to the cosine, $$ 2\cos4x+8\cos3x+8\cos2x-8\cos x-10 $$ so the integral is $$ -\frac{1}{16}\int( 2\cos4x+8\cos3x+8\cos2x-8\cos x-10 )\,dx $$ and we get $$ -\frac{1}{32}\sin 4x-\frac{1}{6}\sin3x-\frac{1}{4}\sin2x+\frac{1}{2}\sin x+\frac{5}{8}x+C $$