Note: This only works when \(x\) is measured in radians. We are now going to look at more complex trigonometric functions where we will use the general rule: \(\int {\cos (ax + b)dx = \frac{1}{a}} ...

The way I would normally do this problem is let u=(cosx), so du=sinxdx Then its just the Integral of u^2? No? With 1/sinx pulled outside of the integral. This is easy to integrate, but apparently is ...