Two ways to use trigonometry in integrals

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

$$ \begin{align} \int sin^3(&x) \, dx \\[5pt] \int sin^2(&x) tan(x) \, dx \\[5pt] \int cos^4(&x) sin^2(x) \, dx \end{align}$$

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,

First some important trig. identities

First we recall the Pythagorean identity:

$$sin^2(x) + cos^2(x) = 1$$

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

$$cos(2\theta) = cos^2(\theta) - sin^2(\theta)$$

we can use the Pythagorean identity to substitute   1 - cos²θ   for sin²θ 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:

Example 1

Odd powers only   $\int sin^3(x) \, dx$

Solution: To do this integral, recognize that sin³x = sin(x)·sin²(x), and write an equivalent integral:

$$\int sin(x) sin^2(x) \, dx$$

Now use the Pythagorean identity

$$sin^2(x) = 1 - cos^2(x)$$

to replace sin²x and rewrite the integral:

$$\int sin(x)(1 - cos^2(x)) \, dx$$

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:

Example 2

$\int tan^3(x) \, dx$

Solution: 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.

$$\int tan(x) sec^2(x) \, dx$$

One substitution is

let u = sec(x),
then du = sec(x)tan(x) dx

Then our new integral can be done very simply:

$$\int u\, du \ \frac{u^2}{2} + C$$

Example 3

A mixed sine-cosine integral:   $\int sin^3(x) cos^2(x) \, dx$

Solution: 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.

$$\int sin(x) (1 - cos^2(x)) cos^2(x) \, dx$$

Expand the integral to get the integral of a sum.

$$\int [sin(x) cos^2(x) - sin(x) cos^4(x)] \, dx$$

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

Example 4

Larger powers   $\int sin^4(x) cos^5(x) \, dx$

Solution: 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

$$\int sin^4(x) cos^5(x) \, dx =$$

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

$$= \int sin^4(x) (1 - sin^2(x))^2 cos(x) \, dx$$

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

$$= \int sin^4(x) [1 - 2 sin^2(x) + sin^4(x)] cos(x) dx$$

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

$$ \begin{align} = \int sin^4(&x) cos(x) \, dx \\ &- 2 \int sin^6(x) cos(x) \, dx \\ &+ \int sin^8(x) cos(x) \, dx \end{align}$$

Example 5

A mixed tan-sec integrand   $\int tan^3(x) sec^2(x) \, dx$

Solution: By now I hope you're getting the hang of these. Focus on the odd power on the tangent by replacing tan²(x) with sec²(x) - 1.

$$\int tan(x) (sec^2(x) - 1) sec^2(x) \, dx$$

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

$$\int [tan(x) sec^4(x) - tan(x) sec^2(x) \, dx$$

Then make the same substitution in both,

Solving integrals of the form   $\int sin^m(x) cos^n(x) dx$   when both m and n are even

Example 6

Evaluate   $\int sin^2(x) \, dx$

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

$$sin^2(x) = \frac{1 - cos(2x)}{2}$$

to make the simple substitution

$$int sin^2(x) \, dx = \frac{1}{2} \int [1 - cos(2x)] \, dx$$

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.

$$= \frac{1}{2} \left[ \int dx - \int cos(2x) \, dx \right]$$

Example 7

Even exponents:   $\int sin^2(x) cos^2(x) \, dx$

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

to get a new integral:

$$ \begin{align} \frac{1}{4} \int (1 - &cos(2x))(1 + cos(2x) \, dx) \\ &= \frac{1}{4} \int (1 - cos^2(2x)) \, dx \end{align}$$

Now this integral contains a cos² term. We can use the power reduction formula again here, this time with 2x substituted for x:

$$cos^2(2x) = \frac{1 + cos(4x)}{2}$$

Practice problems

Find the following indefinite integrals:

Video examples

Example 1

Itegration of tan(x)·cos²(x) dx. Requires integration by parts.

Minutes of your life: 2:38