How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta $? How to solve $\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4 d\theta  $?
I have tried substitution of cosine and sine, dividing and multiplying by Sine and cosine, converting everything to tan. It can probably be solved using beta/gamma function but I can't figure it out.
 A: Let $\theta =2t $ and use the identities $1+\cos2t=2 \cos^2 t,\>
\sin2t = 2\sin t \cos t$
to rewrite the integral as
$$I=\int_0^\pi \sin ^2 \theta(1 + \cos \theta)^4d\theta = 128\int_0^{\pi/2} (\cos^{10}t - \cos^{12} t )dt$$
Then, apply the recursive result
\begin{align}
K_n= \int_0^{\pi/2} \cos^ntdt = \frac{n-1}n K_{n-2},\>\>\>\>\>K_0=\frac\pi2
\end{align}
to arrive at
$$I=128 (K_{10}- K_{12} )=\frac{32}3K_{10}=\frac{32}3\frac9{10}\frac78\frac56\frac34\frac12K_0=\frac{21}{16}\pi$$
A: There is another solution using the "reverse" of multiple angle formulae. In your case
$$(1-\cos^2(t))(1+\cos(t))^4=$$ $$\frac{3 \cos (t)}{2}-\frac{15 \cos (2 t)}{32}-\frac{5\cos (3 t)}{4} -\frac{13\cos (4 t)}{16}
   -\frac{\cos (5 t)}{4} -\frac{\cos (6 t)}{32} +\frac{21}{16}$$
Have a look here.
A: The integral is equal to $$\frac12\int_{-\pi}^\pi\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^2\left(1+\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^4d\theta=-\frac{1}{128}\int_{-\pi}^\pi e^{-6i\theta}(e^{2i\theta}-1)^2(e^{i\theta}+1)^8\,d\theta,$$ and if we imagine the integrand written as a linear combination of exponentials, then all the exponentials vanish after the integration, and only its constant term "survives". It is equal to $$[z^6](z^2-1)^2(z+1)^8=\binom{8}{2}-2\binom{8}{4}+\binom{8}{6}=28-2\cdot70+28=-84,$$ so the integral is equal to $\dfrac{84}{128}\cdot2\pi=\dfrac{21}{16}\pi$.
A: Idea: use $\sin^2\theta = 1- \cos^2 \theta$ and work out the polynomial in $\cos \theta$, and apply techniques from here, e.g., or here on this site in a more modern look.
