Powers of sinus and cosinus are integrated in the following way:

**Odd powers**$$\int \sin^{2n+1}(x)\,\mathrm{d}x=\int \sin^{2n}(x)\sin x\,\mathrm{d}x=\int (1-\cos^2(x))^n\sin x\,\mathrm{d}x$$Expanding the binomial $(1-\cos^2(x))^n$ results in several integrals of the form

$$\int \cos^m(x)\sin x\,\mathrm{d}x$$all of which are solvable with the change of variable $t=\cos x$, $\mathrm{d}t=-\sin x \,\mathrm{d}x$.

The integral

$$\int \cos^{2n+1}(x)\,\mathrm{d}x=\int \cos^{2n}(x)\cos x\,\mathrm{d}x=\int (1-\sin^2(x))^n\cos x\,\mathrm{d}x$$is solved in an analogous manner

**Even powers**$$\int \sin^{2n}(x)\,\mathrm{d}x=\int (\sin^2(x))^n\,\mathrm{d}x=\int \left(\dfrac{1-\cos(2x)}{2}\right)^n\,\mathrm{d}x$$$$\int \cos^{2n}(x)\,\mathrm{d}x=\int (\cos^2(x))^n\,\mathrm{d}x=\int \left(\dfrac{1+\cos(2x)}{2}\right)^n\,\mathrm{d}x$$and after expanding the binomial, the integrand is expressed in terms of lower powers of $\cos$; this way we may proceed iterating until everything is integrated