xaktly | Calculus

Trigonometric integrals

Trigonometric integrals span two sections, this one on integrals containing only trigonometric functions, and another on integration of specific functions by substitution of variables for trig. functions. Both are useful, so make sure to check out the second, too.

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,

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,

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


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:

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

$$sec^2(\theta) = 1 + tan^2(\theta)$$

(obtained by dividing the Pythagorean identity through by cos2θ), to obtain the tangent power-reduction formula:

$$tan^2(\theta) = 1 - \frac{2 tan(\theta)}{tan(2 \theta)}$$

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.


Example 1

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


Solution: To do this integral, recognize that sin3x = sin(x)┬Ěsin2(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:

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).

$$\int sin^3(x) \, dx = \frac{1}{3} cos^3 (x) - cos (x) + C$$

We were able to do the integral above with simple substitutions using the Pythagorean identity 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.


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:

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

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

$$\int tan(x) sec^2(x) \, dx = \frac{1}{2} sec^2(x) + C$$

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

$$\int tan(x) \, dx = \int \frac{sin(x)}{cos(x)} \, dx$$

Then make the substitution

let u = cos(x),
then du = -sin(x) dx

That gives us a new integral that's easy to solve:

$$-\int \frac{du}{u} = -|ln|u| + C$$

Back-substituting for u gives

$$\int tan(x) \, dx = -ln|cos(x)| + C$$

Putting both parts together gives us the solution:

$$\int tan^3(x) \, dx = \frac{1}{2} sec^2(x) + ln|cos(x)| + C$$


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:

let u = cos(x), then du = -sin(x) dx

to get these integrals:

$$-\int u^2 \, du + \int u^4 \, du$$

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

$$-\frac{1}{3} cos^3(x) + \frac{1}{5} cos^5(x) + C$$

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


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.

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

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

let u = sin(x), then du = cos(x) dx

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

$$ \begin{align} &\int u^4 \, du - 2 \int u^6 \, du + \int u^8 \, du \\ \\ &= \frac{1}{5} u^5 = \frac{2}{7} u^7 + \frac{1}{9} u^9 + C \end{align}$$

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

$$ \begin{align} \int &sin^4(x) cos^5(x) \, dx \\ &= \frac{1}{5} sin^5(x) - \frac{2}{7} sin^7(x) + \frac{1}{9} sin^9(x) + C \end{align}$$

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


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,

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

and solve the easy power integrals:

$$\int u^3 \, du - \int u \, du = \frac{1}{4} u^4 - \frac{1}{2} u^2 + C$$

Finally, back substitute to get the solution.

$$ \begin{align} \int tan^3(x) &sec^2(x) \, dx \\ &= \frac{1}{4} sec^4(x) - \frac{1}{2} sec^2(x) + C \end{align}$$



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.

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

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

giving the integral

$$\frac{x}{2} - \frac{1}{4} \int cos(u) \, du$$

We then take the result,

$$= \frac{x}{2} - \frac{1}{4} sin(u) + C$$

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

$$\int sin^2(x) \, dx = \frac{x}{2} - \frac{1}{4} sin(2x) + C$$


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:

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

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

$$\int sin^2(x) cos^2(x) dx = \frac{x}{8} + \frac{1}{32} sin(4x) + C$$

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


Practice problems

Find the following indefinite integrals:

1. $$\int sin^4(x) \, dx$$
2. $$\int cos^2(x) tan^3(x) \, dx$$
3. $$\int sin^2(x) cos^4(x) \, dx$$

4. $$\int tan^3(x) \, dx$$
5. $$\int cos^2(x) tan^3(x) sec^2(x) \, dx$$
6. $$\int sin^5(x) cos^2(x) \, dx$$


Video examples


Example 1

Itegration of tan(x)┬Ěcos2(x) dx. Requires integration by parts.

Minutes of your life: 2:38


Example 2

...

Minutes of your life: 0:00



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