Solving the integral of $sin^n(x)$ I want to delete this question. The results I had trouble with while proving them are false. Furthermore, the question was marked as duplicate by (at least) five users.
For odd $n$ you can write the integral as

$$\int \sin^{2k+1} x\; \mathrm dx = \int \sin^{2k} x \sin x\;\mathrm dx = - \int (1-t^2)^{k} \;\mathrm dt$$ and then use change of variable $t = \cos x$
For even $n=2k\geq2$ use integration by part to write
$$
I=\int\sin^{2k}x\,dx=-\sin^{2k-1}x\cos x+(2k-1)\int\sin^{2k-2}\cos^2x\,dx
$$
and the basic trig relation $\sin^2x+\cos^2x=1$ allows to write the last expression as
$$
I=\sin^{2k-1}x\cos x-(2k-1)\int\sin^{2k}x\,dx+(2k-1)\int\sin^{2k-2}x\,dx
$$
i.e.
$$
I=\frac1{2k}\left[\sin^{2k-1}x\cos x+(2k-1)\int\sin^{2k-2}x\,dx\right]
$$
Repeat the procedure $k-1$ more times.