We use trigonometric functions quite often in integration, even when there are no trig. functions explicitly in the integral. In *this* section we'll look at integrals that contain only trigonometric functions, such as

Trigonometric integrals like these are very important throughout science and engineering. The field of **Fourier analysis**, which includes **fourier series** and the "**fourier transform**", is based upon series of terms that include trig. functions,

and we often have to integrate complicated trigonometric expressions. You couldn't get an MRI image of your aching knee, for example, if it weren't for the Fourier transform and trigonometric integrals.

We'll develop some tricks for making these complicated integrals possible. This section will build heavily on your knowledge of analytical trigonometry (trig. identities), integration by u-substitution and integration by parts.

In another section, we'll cover how it's possible to integrate some pretty tricky functions by substituting trig. functions for variables, taking advantage of some of the special algebra of trig functions, such as the Pythagorean identity,

First we recall the **Pythagorean identity**:

.

If we begin with the cosine double angle formula (derived here),

we can use the Pythagorean identity to substitute 1 - cos^{2}θ for sin^{2}θ to obtain one of the **power-reduction identities**:

Notice that this identity allows us down-convert the power of the cosine function from 2 to 1. And it's easy to integrate a function like cos(2θ) or sin(2θ) by simple u-substitution.

This will be very valuable in doing trig. integrals. Likewise, we can develop a **sine-power conversion formula**:

In a similar way, we can take the Pythagorean identity for secant and tangent,

(obtained by dividing the Pythagorean identity through by cos^{2}θ), to obtain the tangent power-reduction formula:

Here is a recap of the most important trig. identities we'll use to do trigonometric integrals:

In the examples that follow, we'll first do integrals in which at least one of the exponents of a trig. function is *odd*. These will turn out only to need the Pythagorean identities. When all exponents of trig. functions are *even*, we'll need to use the power-reduction formulae to *create* an odd exponent, then proceed as before.

To do this integral, recognize that **sin ^{3}x = sin(x)·sin^{2}(x)**, and write an equivalent integral:

Now use the Pythagorean identity

to replace **sin ^{2}x** and rewrite the integral:

Now if we distribute the **sin(x)** into the parenthesis, this new integral is a sum of two integrals, the last of which can be evaluated easily using the substitution **u = cos(x)**, like this:

The first integral is easy, it's just **-cos(x)**. The second is relatively easy because of the substitution.

Now we just back substitute **cos(x)** for **u** to get the solution (don't forget the constant).

*because* the coefficient of the sine function was **odd**. In order for a substitution like this to work in a trigonometric integral, at least one of the exponents of a trig. function must be **odd**.

This is an example very similar to the first, but it will be a little more complicated to integrate. First use the Pythagorean identity for tangent and secant

From this, multiplying through by tan(x), we get two integrals (We'll call them **1** and **2**), each of which can be solved by simple substitution (u-substitution). Here's what it looks like:

**Integral 1**: The first integral suggests a substitution because tan(x) and sec(x) are related by a derivative.

One substitution is

Then our new integral can be done very simply:

Re-substitution of sec(x) for u gives:

**Integral 2**: The second integral is one you might just remember, but in case you've forgotten, expand tangent into sine/cosine:

Then make the sub

The

The

The

∫ sin

Using the Pythagorean identity, we can take advantage of the odd power of the sine part to reduce it to **sin(x)**, which will come in handy later.

Expand the integral to get the integral of a sum.

Each of these is now easy to do with the *same* simple substitution:

to get these integrals:

Evaluation of these and re-substitution of **cos(x)** for **u** gives the final integral:

.

The only thing that remains is to do an example in which the powers of the trig. functions are higher.

This integral is clearly more complicated simply because of the large powers of the trig. functions. But the fact that one of them is odd presents us with the handle we need to integrate it by hand

We use the Pythagorean identity for sine and cosine to re-express cos4(x) in terms of sines:

Expanding the binomial (1 - sin2(x))2 gives:

and multiplying through by sin4(x) gives us three separate terms which can be integrated separately.

But now each of our terms contains cos(x), whcih we can make use of in a u-substitution:

Notice that cos(x) dx is in each of our integrals, so the substitution gives integrals that are easy to evaluate:

Finally, we can substitute sin(x) back in for u to get the result:

Just so we don't leave the tangent function out, we'll do another example below with a mixed tangent-secant integrand.

∫ tan

By now I hope you're getting the hang of these. Focus on the odd power on the tangent by replacing **tan ^{2}(x)** with

Now expand to get the integral of a sum (which is the sum of integrals):

Then make the same substitution in both,

and solve the easy power integrals:

Finally, back substitute to get the solution.

For integrals with only even powers of trigonometric functions, we use the power-reduction formulae,

,

to make the simple substitution

Then we can separate this integral of a sum into the sum of integrals. The first is trivial, and the second can be don by u-substitution.

Both integrals are easy now (the first is already done below). The second can be done with the substitution

giving the integral

We then take the result,

,

and back substitute **u = 2x** to arrive at the solution to the integral:

∫ sin

First we use our sine and cosine power-reduction formulas in this simple substitution:

to get a new integral:

Now this integral contains a **cos ^{2}** term. We can use the power reduction formula again here, this time with

Make the substitution and carry out the integration. To integrate the last term, we use the u-substitution **u = 4x**, then **du = 4dx**. Be careful with all of the various fractional factors that result.

Finally, the integral is

That should be enough examples for you to try some on your own.

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