# Trigonometric integrals; choosing which one to break up?

1. May 6, 2014

### JessicaJ283782

trigonometric integrals; choosing which one to "break up?"

When you have two different trigonometric functions multiplied together within the integral, for example integral of (cos^4*sin^6) how do you tell which one to "break them up" to substitute an identity in?

Thank you!

2. May 6, 2014

### mathman

I doubt if there is a general rule. In the above cos^4 = (1-sin^2)^2.

3. May 6, 2014

### JessicaJ283782

Do you have any hints on how to pick which one by any chance? Or how you can look at them and tell which one to "break up"?

4. May 6, 2014

### eigenperson

For integrals involving products of cosines and sines, the general principle is as follows:

1. If one of the trig functions has an odd exponent, substitute for the other function. If they are both odd, substitute for either one -- it's your choice. (Example: if the problem is to integrate $\cos^3 x \sin^4 x$, cosine has an odd exponent, so put $u = \sin x, du = \cos x\,dx$.) You will have to use the identity $\cos^2 + \sin^2 = 1$ to carry out the substitution. [Note: This even works if one of the functions is not there at all! If you have to integrate $\sin^3 x$, that is equal to $\sin^3 x \cos^0 x$. Sine has an odd exponent, so substitute for $\cos x$. It will work out!]

2. If both of the trig functions have even exponents, sigh, then use the identities $\cos^2 x = {1 + \cos 2x \over 2}$ and $\sin^2 x = {1 - \cos 2x \over 2}$ until everything is in terms of $\cos 2x$. Then substitute for 2x. Now you have a new problem where the exponents are lower. You may repeat step 1 or 2 as appropriate until the problem is solved.

5. May 10, 2014

### HallsofIvy

You try one and see if it works!