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

- Trigonometric integrals and
- Integration by substitution of trig. functions (or "trig. substitution").

Working with just a few key trigonometric identities can open up a whole range of trigonometric integrals for you. For example, it's very important to be able to do integrals like
### Important trig. identities

because they arise in an important application of mathematics called Fourier Analysis. Fourier integrals are important in physics, chemistry, medical imaging, communications, noise filtering and many other fields.

You couldn't get an MRI (magnetic resonance imaging) for your aching knee if we didn't have Fourier analysis.

First we recall the **Pythagorean identity**:

.

If we begin with the cosine double angle formula,

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 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:

One use of **trigonometric identities** in integrals is to reduce the power (exponent) of one part of the integrand by one.

← Here are the main identities to keep in mind when doing integrals of trigonometric functions.

Now we'll use them to solve integrals of the type

These will be divided into two categories below, those where at least one of m or n are **odd**, and integrals where both m or n are **even**. The best way to do this is by example, so here we go ...

To do this integral, regognize that **sin**^{3}x = sin(x)·sin^{2}(x), and write the new integral:

Now use the identity

to replace **sin ^{2}x** and write the new integral

Now 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 easy because of the substitution.

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

We were able to do this integral with just simple substitution of trig. identities *because* the coefficient of the sine function was odd. In order for simple substitution like this to work, at least one of the exponents of the trig. functions has to 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 we get two integrals, each of which can be solved by simple substitution (u-substitution). Here's what it looks like:

Putting it all together (and remembering that the integral on the right had a minus sign in front of it), we get the result:

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 this simple substitution:

to get these integrals:

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

.

When an **odd power** of a trig. function is present in an integrand, the object is to convert it to a product containing the** first power** of that function. That first power will come in handy later when integrating by u-substitution.

First substitute **(1 - sin ^{2}x)^{2}** for

Now continue to expand the integrand (I'm a poet and I don't even realize it!), to get three integrals:

These can all be solved using the u-substitution, Let u = sin(x) & du = cos(x) dx to get the string of easy to solve polynomial integrals:

Then back-substituting **sin(x)** for **u**, we arrive at the solution:

You can differentiate this result to see that it gets you right back to the original integrand.

By now I hope you're getting the hang of these. Go for the odd power on the tangent by replacing **tan**^{2}(x) with **sec**^{2}(x) - 1.

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

Then make a substitution,

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).

We then take the result,

,

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

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

Why do we need to learn to do trig. integrals like this by hand? One important reason is that they are crucial in understanding the **Fourier transform**, something you might encounter later in math. The **FT** is extremely important in many aspects of modern life. You couldn't get an **MRI** on a sore joint without the Fourier transform, a way of transforming between the time and frequency domains in any study of electromagnetic radiation. There are other uses for trig. integrals, including the study of noise reduction and signal processing (like sending data), but the FT is a big one.

So far we've solved trigonometric integrals using trig. identity substitution and a few other small tricks. Now let's substitute some trigonometric functions for algebraic variables in algebraic expressions like these (**a** is a constant):

This will be a nifty trick to solve some integrals you probably haven't cracked yet.

Let's start with the first radical expression, **(a**^{2} - x^{2})^{1/2}. We begin by introducing a dummy variable, **t**, and making the substitution **x = a sin(t)**. Then

The substitution has reduced a radical to a simple trigonometric expression, the integral of which we know, so there's hope for this kind of substitution.

In a similar way we can substitute **x = a tan(t)** for the **x** in the second radical and **x = a sec(t)** for the **x** in the third. Each substitution leads to a simple trigonometric function.

See the table below for a summary of integration by trig. substitution.

To begin, consult the table above and make the substitution **x = a sin(t)**, where **a = 9** (the square root of 81):

Substitute and simplify the expression under the radical.

That last step uses the Pythagorean identity, **sin ^{2}t + cos^{2}t = 1**. Now the integrand, including dx, can be written:

The substitution allows us to take the square root:

Now move the constant out and use our cosine power-reduction formula to change the cos^{2}(t) term into something we can integrate, just as in the examples above. Integrate the cos(2t) integrand with simple u-substitution.

The integral in terms of the dummy variable * t* is now easy to write, and we can substitute expressions containing

Now let's go back to the definition of t that we started with, namely that **sin(t) = x/9**, and construct the right triangle (angles not to scale) that goes with it. Note that we use the Pythagorean theorem to get the length of the lower side of the triangle.

That gives us **cos(t)**, so we can plug **sin(t)** and **cos(t)** into the previous result to get

Now just some easy cancellation gives us the final integral:

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