Trigonometric integrals; choosing which one to break up?

[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!

 

[mathman]

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!

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

 

[JessicaJ283782]

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

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"?

 

[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.

 

[HallsofIvy]

You try one and see if it works!