## Trigonometric Integrals

A trigonometric integral is an integral involving products and powers of trigonometric functions: cosine, sine, tangent, secant, cosecant, and cotangent. Many of these integrals can be handled with u-substitution, but there are other methods which are outlined in this module. The three families of integrals discussed in this module are

## Product of sines and cosines

Consider the integral

There are several cases to consider based on whether and are odd and even.

### m is odd

If is odd, then one factor of can be set aside. This leaves behind an even power of , which can be expressed in terms of using the Pythagorean identity. Then the substitution can be made.

**Example**

Find

Following the above outline, set aside one factor of , which gives

Now, there is an even power of sine remaining, which can be rewritten using the Pythagorean identity

This gives

Now, the integral can be handled by letting (and ).

### n is odd

If is odd, the procedure is very similar. This time, we set aside a factor of . This leaves an even power of which can be expressed in terms of using the Pythagorean identities.

**Example**

Find

Following the procedure outlined above, we set aside a factor of cosine and use the Pythagorean identity, which gives

Now, we are ready to make the substitution , . This gives

### Both m and n are even

If neither nor is odd, then both are even. This is a bit more difficult and requires using the power reduction formulas:

Power reduction |
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**Example**

Find

Using the first power reduction formula gives

**Example**

Find

Using both the power reduction formulas and doing some algebra gives

**Example**

Find

Using the second power reduction formula (and then again in a later step) gives

## Product of tangents and secants

Next consider the integral

As with the product of sines and cosines, the method will depend on whether and are odd or even.

### m is odd

If is odd, we will set aside a factor of . Note that this is the derivative of and so this sets up a substitution of . After setting aside these factors, we are left with an even power of , which can be expressed in terms of using the Pythagorean identity

Now, the integral can be computed using the substitution .

**Example**

Compute

Since the power of tangent is odd, we set aside a factor of , and use the Pythagorean identity to find

Now, we can make the substitution

which gives

### n is even

If is even, then we can set aside a factor of . Note that this is the derivative of and therefore sets up the substitution . Setting aside leaves an even power of , which can be expressed in terms of using the Pythagorean identity

Then the substitution allows the computation of the integral.

**Example**

Compute

Because the power of secant is even, we set aside and use the Pythagorean identity to find

Now, we are prepared for a substitution of

Making this substitution and simplifying gives

### m is even, n is odd

If neither of the above cases holds, then the integral is a bit more difficult. It typically requires a bit of algebra and several applications of a reduction formula (or integration by parts). A general method is to rewrite the even power of tangent entirely in terms of secant by using the Pythagorean identity

This gives an integral which is sums of powers of . Each of these can be solved using the reduction formula for secant:

along with the fact that

**Example**

Compute

Using the Pythagorean identity gives

Now, using the reduction formula on the first of these integrals gives

Combining this with the above expression and using the integral of secant, we find

## Product of sine and cosine with constants

Finally, consider the integral

This integral requires some algebra to simplify the integrand. One can verify using the sum and difference formulas for sine that

This expression can be integrated term by term to find

There are similar formulas for related integrals:

These formulas need not be memorized, but be aware they exist and look them up when necessary.

**Example**

Compute

Using the formula given, we have

where we have used the fact that cosine is even to simplify the final expression.

## Additional examples

**Example**

Compute

Use the fact (which is proven using integration by parts) that

Applying the limits of integration in the above reduction formula gives

Now, notice that

because cosine is 0 at . Therefore,

This is itself a reduction formula. By computing the base cases and , respectively, we find

and

Now the value of the integral for any higher value of can be found by repeatedly using the above formula until the integral reduces to one of the base cases above. Using induction, one finds

**Example**

Compute

Here the power of tangent is odd, so the method calls for setting aside a factor of . However, there is no factor of secant in this integral! It turns out that this is not a problem; we can multiply the top and the bottom by secant to introduce a factor of secant, and the algebra works out:

Now, proceed as usual with the substitution

Thus,

## EXERCISES

Compute the following indefinite integrals. You may need to use reduction formulae or coordinate changes.