For "trigonometric powers" integrals where the two functions are "linked" by the Pythagorean Identity, you want to try to arrange for one of the factors, which is a power of one trig function, to be written in terms of an even power of the other trig function. So for powers of sine and cosine in the integral $ \ \int \ \sin^m x \ \cos^n x \ \ dx \ $ , we have
for $ \ m \ $ odd and $ \ n \ $ even -- extract a factor of $ \ \sin x \ $ to place with the differential $ \ dx \ $ , writing $ \ \int \ \sin^{m-1} x \ \cos^n x \ \ ( \sin x \ dx) \ $ , then make the substitution $ \ du \ = \ \sin \ x \ dx \ \Rightarrow \ u \ = \ -\cos \ x \ $ ; with $ \ m - 1 \ = \ 2p \ $ , we have
$$ \ \int \ \sin^{2p} x \ \cos^n x \ \ ( \sin x \ dx) \ \ \Rightarrow \ \ \int \ (1 - \cos^2 \ x)^p \ \cos^n x \ \ ( \sin x \ dx) $$
$$ \Rightarrow \ \ \int \ (1 - u^2 )^p \ (-u)^n \ \ du \ \ ; $$
for $ \ m \ $ even and $ \ n \ $ odd -- extract a factor of $ \ \cos x \ $ , write $ \ \int \ \sin^m x \ \cos^{n-1} x \ \ ( \cos x \ dx) \ $ , $ \ du \ = \ \cos \ x \ dx \ \Rightarrow \ u \ = \ \sin \ x \ $ ; with $ \ n - 1 \ = \ 2q \ $ , we have
$$ \ \int \ \sin^m x \ \cos^{2q} x \ \ ( \cos x \ dx) \ \ \Rightarrow \ \ \int \ \sin^m x \ (1 - \sin^2 \ x)^q \ \ ( \cos x \ dx) $$
$$ \Rightarrow \ \ \int \ u^m \ (1 - u^2 )^q \ \ du \ \ ; $$
for $ \ m \ $ and $ \ n \ $ both odd -- use either of the above;
for $ \ m \ $ and $ \ n \ $ both even-- we use the "sine/cosine-squared identities", $ \ \sin^2 \ x \ = \ \frac{1}{2} \ (1 \ - \ \cos \ 2x) \ $ $ \ \cos^2 \ x \ = \ \frac{1}{2} \ (1 \ + \ \cos \ 2x) \ $ ; with $ \ m \ = \ 2p \ $ and $ \ n \ = \ 2q \ $ , we have
$$ \ \int \ \sin^{2p} x \ \cos^{2q} x \ \ dx \ \ \Rightarrow \ \ \int \ \left[ \ \frac{1}{2} \ (1 \ - \ \cos \ 2x) \ \right]^p \ \left[ \ \frac{1}{2} \ (1 \ + \ \cos \ 2x) \ \right]^q \ \ dx \ \ . $$
[Special case: if $ \ m \ = \ n \ $ , odd or even , we can just use
$$ \ \int \ \sin^m x \ \cos^m x \ \ dx \ \ \Rightarrow \ \ \int \ ( \ \sin \ x \ \cos \ x \ )^m \ \ dx \ = \ \int \ \left( \ \frac{1}{2} \ \sin \ 2x \ \right)^m \ \ dx \ \ . \ ] $$
You will unavoidably have some sort of binomial expansion to carry out if the smaller power is larger than three. There are analogous methods for powers of tangent and secant, and for powers of cotangent and cosecant.