Evaluate $\int{\sin^{13}x}dx$ Evaluate $\int \sin^{13}x\,dx$.
My solution so far:
$$\int\sin^{13}x\,dx=\int\sin x\sin^{12}x\,dx=\int\sin x(\sin^{2}x)^6\,dx$$
Let $t=\cos x$, then $dt=-\sin x\,dx$.
Now we have: $$\int\sin^{13}x\,dx=-\int(1-t^2)^6,\ dt$$.
How do I proceed from here? I need detailed answer.

\begin{align}
\int(1-t^2)^6 &= \int \sum_{n=0}^6 {6 \choose n} (-t^2)^n \\
&= \sum_{n=0}^6 {6 \choose n} (-1)^n \int t^{2n} \\
&= \sum_{n=0}^6 {6 \choose n} (-1)^n \left( \frac{t^{2n+1}}{2n+1}\right) + C \\
&= \frac{t^{13}}{13} - \frac{6 t^{11}}{11} + \frac{5t^9}{3} - \frac{20t^7}{7} + 3 t^5 - 2t^3 + t + C
\end{align}
