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

In this section, we will study how to integrate certain types of trigonometric functions. We begin by giving the antiderivatives of the six basic trigonometric functions:

### Antiderivatives of other trigonometric functions

We are able to establish the antiderivatives of $\sin (x)$ and $\cos (x)$ easily from the formulas for their derivatives, but how do we establish the other formulas?

The formulas for the antiderivatives of $\csc (x)$ and $\cot (x)$ are found analogously.

### Integrals involving products of $\sin (x)$ and $\cos (x)$

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 $u=x^2$, $\d u = 2x \d x$, so the $2x \d x$ in the original integral is part of the new differential $\d u$ 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 $u=x^2$, $\d u = 2x \d x$, so the $2x \d x$ in the original integral is part of the new differential $\d u$ and:

where we have used the fact that $u=x^2$ 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.

We can make a useful observation from this example. Our previous integral involved powers of $\sin (x)$ and $\cos (x)$. We peeled off a copy of the derivative of what we wanted to use for $u$. 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 $u$.

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

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
: $\cos ^2(\theta )= \frac {1}{2}+\frac {1}{2}\cos (2\theta )$
Sine Power-Reduction
: $\sin ^2(\theta ) = \frac {1}{2}-\frac {1}{2}\cos (2\theta )$
Using the above formula we find $\sin ^2(2x) = \frac {1}{2}-\frac {1}{2} \cos (\answer {4x})$.

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:

#### Working with products of powers of $\sec (x)$ and $\tan (x)$

The same logic used in the preceding examples can be used to find antiderivatives involving products of $\sec (x)$ and $\tan (x)$. Indeed, the Pythagorean identity has two other forms: where the first identity we find by dividing the Pythagorean identity by $\cos ^2(x)$ 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.

There is no analogue for the power reduction formulas for sines and cosines, so sometimes, these integrals can become a bit trickier.

This last example can be generalized to a reduction formula for $\int \sec ^{n}(x) \d x$. We omit the details but using integration by parts in a similar fashion as above with $u=\sec ^{n-2}(x)$ and $\d v=\sec ^{2}(x)$ 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 complimentary trigonometric functions (sines and cosines, tangents and secants, or cosecants and cotangents), we can employ a general strategy to find the antiderivatives:

• Let $u$ 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 $u$.
• 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