For trig integrals involving powers of sines and cosines, there are two important cases:

1. The integral contains an odd power of sine or cosine.

2. The integral contains only even powers of sines and cosines.

I will look at the odd power case first. It turns out that the same idea can be used to integrate some powers of secants and tangents, so I'll digress to do some examples of those as well.

* Example.* Compute .

In this example, the key point was in the second line. I obtained an integral with lots of 's and a single . This allowed me to make the substitution , because the was available to make .

I got the by "pulling it off" the odd power of . Then I converted the rest of the stuff to 's using the identity . This is the generic procedure when you have at least one odd power of sine or cosine.

* Example.* Compute .

If you have an integral involving sines and cosines in which all the
powers are *even*, the method I just described usually won't
work. Instead, it is better to apply the following *
double angle formulas*:

Any even power of or can be expressed as
a power of or . Use the
identities above to substitute for or , and multiply out the result. The net effect is to
* reduce the powers* that occur in the integral,
while at the same time increasing the arguments ( ).

* Example.* Compute .

* Example.* Compute .

I'll use the double angle formula (twice):

* Example.* Why would it be a bad idea to use the
double angle formulas to compute ?

Suppose I try to apply the double angle formula for cosine:

The integral can be done in this form, but you either need to apply
one of the angle addition formulas to or use
integration by parts. The problem is that *having trig functions
with different arguments in the same integral makes the integral a
bit harder to do*.

It would have been better to do the integral by using the "odd power" technique:

In some cases, you can use trig identities to do integrals involving sine and cosine. For example, the angle addition and subtraction formulas for cosine are

Add the two equations --- the terms cancel --- and divide by 2:

The formula is

If instead you subtract the equation from the equation and divide by 2, you obtain

Likewise, the angle addition and subtraction formulas for sine are

Adding the equations and dividing by 2 gives

To summarize:

(a) .

(b) .

(c) .

You can use these identities to integrate products of sines and cosines with different arguments. (Note that you can also do these integrals using integration by parts.)

* Example.* Compute .

Using the formula for with and , I get

Note: If you wind up with a "negative angle" in applying the identities, you can get rid of it using the identities

To integrate some powers of secants and tangents, here are two useful approaches:

1. Use to convert the integrand to something with lots of 's and a single . Then substitute .

2. Use to convert the integrand to something with lots of 's and a single . Then substitute .

* Example.* Compute .

In this example, I pulled off a , then converted the rest of the stuff to 's using . The was exactly what I needed to make for the substitution .

Notice that the argument did not play an important role in the problem.

* Example.* Compute .

In this example, I pulled off a , then converted the rest of the stuff to 's using . The was exactly what I nneded to make for the substitution .

* Example.* Compute .

I can do the first integral using , so and :

I can do the second integral using , so and :

Therefore,

* Example.* Compute .

In this problem, I'll use the identity

Applying this to the top of the fraction, I get

I used the formula

If you didn't know this, you could derive it by writing . Then substitute .

I used to do .

* Example.* Compute .

This integral uses a trick:

* Example.* Compute .

I can compute using parts:

Thus,

This integral also comes up a lot, so you should make a note of it.

* Remark.* Using the methods of the last two
examples, you can show:

In general, integrals involving powers of cosecant and cotangent use the same ideas as integrals involving powers of secant and tangent.

* Example.* Compute .

Remember the trig identity

So

The examples show that certain patterns that arise in trig integrals are good, in the sense that they allow you to do a substitution which makes the integral easy. Here is a review of some of the "good patterns":

(a) Lots of 's and a single .

(b) Lots of 's and a single .

(c) Lots of 's and a single .

(d) Lots of 's and a single .

(e) Lots of 's and a single .

(f) Lots of 's and a single .

You should aim for these patterns whenever possible.

Finally, I'll note that you can sometimes use integration by parts to
obtain * recursion formulas* which reduce the
integral of a power of a trig function to the integral of a smaller
power.

* Example.* Derive a recursion formula for for .

Integrate by parts:

After integrating by parts, I used the identity . I multiplied out the terms in the integral, then broke the integral up into two integrals.

Next, add to both sides of the equation:

The last equation is the recursion formula. It reduces the integral of a power of sine to some stuff ( ) plus the integral of a power of sine that is smaller by 2.

Here's how the formula would apply if :

Copyright 2019 by Bruce Ikenaga