We can use substitution and trigonometric identities to find antiderivatives of certain types of trigonometric functions.

### Antiderivatives of other trigonometric functions

We are able to establish the antiderivatives of and easily from the formulas for their derivatives, but how do we establish the other formulas?

The formulas for the antiderivatives of and are found analogously.

### Integrals involving products of and

We begin with a simple example:

In this example, we saw that a substitution was quite helpful. Earlier, we saw that there were two common instances in which substitutions could be helpful to compute antiderivatives:

- Once we change variables, the remaining expressions in the old variable
are part of the differential in the new one. For instance, consider:
By setting , , so the in the original integral is part of the new differential and:

- Once we change variables, it is not complicated to write the remaining old
variables in terms of the new one. For instance, consider:
By setting , , so the in the original integral is part of the new differential and:

where we have used the fact that in the last step above. Note that we can do this last integral using integration by parts!

This highlights a strategy that we will employ in this section; try to let one of the trigonometric functions be our new variable, and express the remaining expressions easily in terms of it.

*even*powers of sines and cosines, so this will be a fruitful approach:

We can use this identity to rewrite in our integral so that it becomes: Now the same strategy we used in the first example of this section will work. We can do the substitution . We find , so that the integral becomes:

Finishing this integral we get

Note that had we let , we will have to reserve a copy of for the differential. This leaves three powers of left! We certainly can use the Pythagorean identity from before to write:

If , we find , and from the above we can write the original integral as:

This integral can be evaluated with another substitution and some algebra, but it is certainly more time-consuming than the first approach!

We can make a useful observation from this example. Our previous integral involved powers of and . We peeled off a copy of the derivative of what we wanted to use for . If this leaves an even number of the other trigonometric function left, we can use the Pythagorean identity to write the remaining powers of the other trigonometric function in terms of .

In the preceding example, the power of cosine was odd, and for other examples, as long as the power of in the original integral is odd, then we can peel off a term to be used for and then let to express the integral in terms of . This general approach is stated in more detail in the exercises, but for now let’s try another example.

If we try , we find that we need a factor of for the differential, which leaves 1 power of left, so using the Pythagorean identity would produce an unpleasant integral. However, if we try , we find that we need a factor of for the differential, which leaves four powers of left. We can rewrite this suggestively as:

Now using the Pythagorean identity we can rewrite everything in terms of , but reserve the one power of to help with the differential in an eventual variable substitution :

Making the substitution

yields This is a polynomial in , and so we can easily finish the integration by just expanding the polynomial out and integrating term-by-term. Write with me Substituting we find our final answer:Sometimes we need to massage our function into the correct form:

### Another strategy

Thinking again about powers of cosine and sine, what do we do when both powers are even? Our strategy above is sunk, since peeling-off one power leaves an odd power, and we cannot use the Pythagorean identity to rewrite this in a nice way. Instead, we will use the “power-reduction formulas”:

- Cosine Power-Reduction
- :
- Sine Power-Reduction
- :

In this case, it is (usually) critical to apply the power-reduction formulas to *every*
instance of cosine and sine appearing in the integrand.

First consider a simple example:

Since this expression has only even powers of sine, we must use the same strategy again, and employ a power-reduction formula: so we find

#### Working with products of powers of and

The same logic used in the preceding examples can be used to find antiderivatives involving products of and . Indeed, the Pythagorean identity has two other forms: where the first identity we find by dividing the Pythagorean identity by and the second we find by subtracting 1 from both sides of the first identity.

The derivatives of these functions also can be expressed in terms of each other:

It is worth seeing several examples where we use these identities. We proceed in a fashion similar to the procedures we saw earlier.

Since the power of secant is even, we can separate off a factor of for our . That leaves an extra factor of that we must deal with. Let’s now write out the details.

Now using the identity we can rewrite everything in terms of , but reserve the one power of to help with the differential in an eventual variable substitution :

Making the substitution

yields This is a polynomial in , and so we can easily finish the integration by just expanding the polynomial out and integrating term-by-term. We leave it to the intrepid young mathematician to finish this problem.Since the power of tangent is odd and the power of secant is odd, we can pull out a factor and we have

Now using the identity we can rewrite the remaining in terms of , but reserve the one power of to help with the differential in an eventual variable substitution :

Making the substitution

yields This is a polynomial in , and so we can easily finish the integration by just expanding the polynomial out and integrating term-by-term. We leave it to the interested mathematician to finish this problem.There is no analogue for the power reduction formulas for sines and cosines, so sometimes, these integrals can become a bit trickier.

Choose

We recall the integration by parts formula , so we obtain

The rightmost integral does not have an odd power of or an even power of but we could convert everything to be in terms of by using the identity .

The rightmost integral then becomes

Combining this with our original integral and using the previous example which gave us we get

Notice that the remaining integral on the right hand side is simply the original integral we were trying to determine. You will remember that we first experienced this phenomenon in the section on integration by parts when we tried to compute . Since the desired unknown quantity appears on the left and the right we simply solve for it and so

This last example can be generalized to a reduction formula for . We omit the details but using integration by parts in a similar fashion as above with and gives us the following reduction formula:

We won’t derive them here but note that there are similar reduction formulas that we could derive for integrating powers of other trigonometric functions using integration by parts.

### Final thoughts

When we encounter integrals that involve products of complementary trigonometric functions (sines and cosines, tangents and secants, or cosecants and cotangents), we can employ a general strategy to find the antiderivatives:

- Let be one of the trigonometric functions. Peel off a copy of the derivative for the differential. If this leaves an even number of the other trigonometric function, use the appropriate Pythagorean identity to write the remaining powers of the other trigonometric function in terms of .
- If the integral involves a product of sines and cosines and the above does not work, try using the power reduction formulas.
- If the integral involves secants and tangents (or cosecants and cotangents), try either integration by parts or convert everything to sines and cosines.

As usual, practice is necessary to develop good intuition when working these types of problems.

“One should study mathematics simply because it helps to arrange one’s ideas” - M. W. Lomonossow