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.