Trigonometric Integrals

This is around capturing intergals in the form of: $\int {\sin }^m(x){\cos }^n(x)dx$.

If m and n are 1, then we can solve this with u-substitution where u=sin(x) or u=cos(x). If m and n are 2, then we can use $co{s}^2(x)+si{n}^2(x)=1$ to figure it out.

If m is odd, you can rewrite it as m=2k+1.

$$\begin{array}{rl}& si{n}^{m}x\\ & \\ & =si{n}^{2k+1}x\\ & \\ & =si{n}^{2k}(x)sin(x)\\ & \\ & =(si{n}^2(x){)}^{k}sin(x)\\ & \\ & =(1-co{s}^2(x){)}^{k}sin(x)\\ & \end{array}$$

This means that for the original formula:

$$\begin{array}{rl}\int & si{n}^m(x)co{s}^n(x)dx\\ & \\ & =((1-co{s}^2(x){)}^{k}sin(x))co{s}^n(x)dx\\ & \\ u& =cos(x)\\ & \\ du& =-sin(x)dx\\ & \\ & =-(1-u^2{)}^k{u}^{n}du\end{array}$$

If n is odd, we don’t have to do the power reduction to convert to cos.

$$\begin{array}{rl}\int & si{n}^m(x)co{s}^n(x)dx\\ & \\ & =si{n}^m(1-si{n}^2(x){)}^{k}cos(x)dx\\ & \\ u& =sin(x)\\ & \\ du& =cos(x)dx\\ & \\ & =u^m(1-u^2{)}^{k}du\end{array}$$

If both are even, then we can reduce the integrand with this:

$$\begin{array}{r}{\cos }^2(x)=\frac{1+\cos (2x)}{2}\\ \\ {\sin }^2(x)=\frac{1-\cos (2x)}{2}\end{array}$$

.. then go back through the process as necessary.