Calculus 2 Practice Final Bonus Question (definite integral) Calculate the definite integral $$\int_0^\pi \sin t · \sin^{11}(\cos t) dt$$
It seems straightforward enough yet seems to be a trick question? Using u-substitution, I took $u=\cos t$ and got $\int_1^{-1} \sin^{11} u du$, which I then split into $\int_1^{-1} \sin^{10} u· \sin u du$. Using a secondary u-sub, I took $a=-\cos u$ to obtain $$\int_1^{-1} {(1-a^2)}^{5} da$$ but now I'm stuck. Expanding this seems much too complicated and a third u-sub doesn't seem to do anything. Should I try approaching this question in a different way?
 A: Starting from the stated material in the proposed problem:
\begin{align}
I &= \int_{0}^{\pi} \sin t \, \sin^{n}(\cos t) \, dt \\
&= \int_{0}^{\pi} \sin^{n}(\cos t) \, d(\cos(t)) \quad \text{let} \, u = \cos t \\
&= \int_{-1}^{1} \sin^{n}(u) \, du \\
&= \int_{-1}^{0} \sin^{n}(u) \, du + \int_{0}^{1} \sin^{n}(u) \, du \quad \text{let} \, u = -x \, \text{in the first integral} \\
&= \int_{0}^{1} \sin^{n}(u) \, du + (-1)^n \, \int_{0}^{1} \sin^{n}(x) \, dx \quad \text{let} x \to u \\
&= (1 + (-1)^n) \, \int_{0}^{1} \sin^{n}(u) \, du \\
&= (1 + (-1)^n) \, \left[ \cos(u) \, {}_{2}F_{1}\left(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2}; \cos^{2}(u) \right) \right]_{0}^{1} \\
&= (1 + (-1)^n) \, \left[\cos(1) \, {}_{2}F_{1}\left(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2}; \cos^{2}(1) \right) - \frac{\sqrt{\pi}}{2} \, \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1 + \frac{n}{2}\right)} \right].
\end{align}
Now, if $n$ is odd then
$$ \int_{0}^{\pi} \sin t \, \sin^{2n+1}(\cos t) \, dt = 0.$$
This applies to the proposed question due to $n$ being $11$. When $n$ is even then
$$ \int_{0}^{\pi} \sin t \, \sin^{2n}(\cos t) \, dt = B_{\cos^{2}(1)}\left(\frac{1}{2}, n + \frac{1}{2}\right) - \frac{\pi}{4^n} \, \binom{2 n}{n}, $$
where $B_{x}(a, b)$ is the incomplete Beta function.
