Integration by Trigonometric Identities

Integrating by using trig identities. I dont know if this is a real technique.

Wrong Example

$\int {\cos }^{3}xdx$ $u=\cos x$ $du=-\sin xdx$ this substitution doesn’t work! there is no $\sin$ term in the original function!

Good Example

$\int {\cos }^{3}xdx$ $\int ({\cos }^{2}x)(\cos x)dx$ $\int (1-{\sin }^{2}x)(\cos x)dx$ let $u=sinx$ $du=\cos xdx$ $\int (1-u^2)dudx=u-\frac{u^3}{3}+c$ $=\sin x-\frac{\sin (3x)}{3}+c$

General Rules

$\int {\sin }^{m}x{\cos }^{n}xdx$

1. If $m$ is odd, convert as many $si{n}^{2}x\to (1-{\cos }^{2}x)$ You’ll be left with $\sin xdx$. Sub $u=cosx$ and $du=-sinxdx$
2. if $n$ is odd, convert as many $co{s}^{2}x\to (1-{\sin }^{2}x)$ You’ll be left with $\cos xdx$ Sub $u=\sin x$ and $du=\cos xdx$
3. If both $n$ and $m$ are even, use Half Angle Identity and rearrange.