
$$\int \sin^n(x) \cos^m(x)\,dx$$
If $n$ is odd or if $m$ is odd, then we can use the identity $\sin^2(x)+\cos^2(x)=1$ to manipulate the integral into the form $\int \sin^k(x) \cos(x)\,dx$ which can be solved with the substitution $u=\sin(x)$; or into the form $\int \sin(x) \cos^k(x) \,dx$ which can be solved with the substitution $u=\cos(x)$.
If $n$ and $m$ are both even, though, this doesn't work and we need to use the double angle formulas $\sin^2(x) = \frac{1}{2} \Bigl(1\cos(2x)\Bigr)$, or $\cos^2(x) = \frac{1}{2}\Bigl(1 + \cos(2x)\Bigr)$ to simplify the problem.
 $$\int \tan^n(x) \sec^m(x)\,dx,$$
If $n$ is odd or $m$ is even we can use the identity $\tan^2(x) + 1 = \sec^2(x)$ to manipulate the integral into the form $\int \tan^k(x) \sec^2(x) \,dx$ which can be solved with the substitution $u=\tan(x)$; or into the form $\int \tan(x) \sec^k(x) \,dx$ which can be solved with the substitution $u=\sec(x)$.
If neither works, then there may be a way to simplify the problem using integration by parts, but it's likely to be hard.
 $$\int \sin(ax)\sin(bx)\,dx$$
There are two methods: one is to integrate by parts twice; the other is to use additionofangle formulas.
