Integrals Involving Trig Functions

Main idea

To integrate powers of sine and cosine, set one factor aside to be the $du$ of a substitution and use $\sin^2 + \cos^2 = 1$ to write everything else in terms of the other function. When no power is odd, there's nothing to set aside, so you lower the powers with the double-angle formulas instead.

The identities you'll use

\sin^2 x = {\htmlData{sym=0}{\tfrac{1 - \cos 2x}{2}}}, \qquad \cos^2 x = {\htmlData{sym=1}{\tfrac{1 + \cos 2x}{2}}}, \qquad \tan^2 x = {\htmlData{sym=2}{\sec^2 x - 1}}

What the symbols mean

$\tfrac{1 - \cos 2x}{2}$
Power reduction for sin²
Swaps a square for a first power of cosine at double the frequency.
$\tfrac{1 + \cos 2x}{2}$
Power reduction for cos²
The same trick for cosine squared.
$\sec^2 x - 1$
Rewrite for tan²
Turns tan² into something you can integrate, since ∫sec² x dx = tan x.

Explanation

A trig integral, in this lesson, is simply an integral of powers of trig functions multiplied together, like $\int \sin^3 x\cos^2 x\,dx$ . They're good for more than practice: averages of AC power, Fourier series and arc lengths all lead to them. All we're really saying is that each one can be untangled into something as simple as $\int u^4\,du$ . Take $\int \sin^4 x\cos x\,dx$ : it's just $\sin^4 x$ (written entirely in terms of $\sin x$ , which we call $u$ ) times $\cos x\,dx$ (exactly $du$ , since the derivative of $\sin x$ is $\cos x$ ), so it's $\int u^4\,du = \frac{\sin^5 x}{5} + C$ . Everything below is about getting other trig integrals into that shape, using $\sin^2 x + \cos^2 x = 1$ to do the rewriting. The cases look like a lot, but it's the same move every time; bear with it.

Sines and cosines, $\int \sin^m x\cos^n x\,dx$ :

Power of sine is odd. Save one $\sin x$ , convert the rest to cosines with $\sin^2x = 1 - \cos^2x$ , and substitute $u = \cos x$ .
Power of cosine is odd. Save one $\cos x$ , convert the rest to sines, and substitute $u = \sin x$ .
Both powers even. Use the power-reduction formulas, as many times as needed.

Why does odd matter? After you save one factor, an odd power leaves an even power behind, like $\sin^2 x$ or $\sin^4 x = (\sin^2 x)^2$ , and even powers are exactly what the identity can convert. The saved factor is the $du$ you need: $d(\cos x) = -\sin x\,dx$ and $d(\sin x) = \cos x\,dx$ . When both powers are even, saving a factor would leave an odd power that won't convert, so you need a different idea. Power reduction trades each square for a first power of cosine at double the frequency, and that you can integrate.

Tangents and secants, $\int \tan^m x\sec^n x\,dx$ , work the same way with a different pair of derivatives: $d(\tan x) = \sec^2 x\,dx$ and $d(\sec x) = \sec x\tan x\,dx$ .

Power of secant is even. Save $\sec^2x$ , convert the rest to tangents with $\sec^2 x = 1 + \tan^2 x$ , and let $u = \tan x$ .
Power of tangent is odd. Save $\sec x\tan x$ , convert the rest to secants, and let $u = \sec x$ .

Two antiderivatives come up so often they're worth knowing by heart:

\int \tan x\,dx = \ln|\sec x| + C, \qquad \int \sec x\,dx = \ln|\sec x + \tan x| + C.

Different frequencies. For $\sin(ax)\cos(bx)$ the save-a-factor trick fails, because the two functions don't share an argument. Instead, rewrite the product as a sum with a product-to-sum identity such as $\sin A\cos B = \tfrac12\big[\sin(A-B) + \sin(A+B)\big]$ , and integrate each term on its own. These integrals are the backbone of Fourier series, as the first picture below shows.

A common mistake. With $\int \sin^2 x\cos^2 x\,dx$ , it's tempting to save a $\cos x$ anyway and write the leftover $\cos x$ as $\sqrt{1 - \sin^2 x}$ . That puts a root in the integral and makes things much worse. Saving a factor only works when what's left over is an even power. When both powers are even, reduce them instead.

See it

This is sin x times sin(kx) over one full period, from 0 to 2π. As you slide k, notice that the net area lands on exactly 0 at k = 2, 3 and 4: the positive and negative lobes cancel perfectly. Only at k = 1 does it come out to π. That's what lets a Fourier series pull each frequency out of a signal without the others getting in the way.

Here sin² x and (1 − cos 2x)/2 are drawn on top of each other, and as you can see they're the same curve. It wobbles between 0 and 1 around an average of ½, twice as fast as sin x. Writing it the second way is what makes it easy to integrate.

Worked examples

Example 1

Find $\displaystyle\int \sin^3 x\cos^2 x\,dx$ .

1
The power of sine is odd, so save one $\sin x$ and turn the remaining $\sin^2 x$ into cosines: $\;\sin^3x\cos^2x = \sin^2 x\cos^2 x\,\sin x = (1 - \cos^2 x)\cos^2 x\,\sin x$ .

Example 2

Find $\displaystyle\int \sin^2 x\,dx$ .

1
There's no odd power to save a factor from, so reduce the power: $\sin^2 x = \tfrac12(1 - \cos 2x)$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $\displaystyle\int \cos^3 x\,dx$ .

Problem 2+5 XP

Evaluate $\displaystyle\int_0^{\pi/2} \sin^4 x\cos^3 x\,dx$ .

Problem 3+5 XP

Find $\displaystyle\int \sin^2 x\cos^2 x\,dx$ .

Problem 4+5 XP

Find $\displaystyle\int \tan^2 x\sec^4 x\,dx$ .

Problem 5+5 XP

Find $\displaystyle\int \cos^2(5x)\,dx$ .

Problem 6+5 XP

A current $I(t) = 10\sin(120\pi t)$ amps runs through a $5$ Ω heater, delivering $P = I^2R = 500\sin^2(120\pi t)$ watts. What's the average power over a cycle?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find $\displaystyle\int\sec^3x\,dx$ .

Where it's used

The average of $\sin^2(\omega t)$ over a full cycle is exactly ½, which is why an AC outlet's "120 volts" means a peak of $120\sqrt2 \approx 170$ volts. Every Fourier series is built from integrals of products of sines and cosines.

Fourier SeriesCalc II · Electrical Engineering RLC CircuitsDiff Eq · Electrical Engineering

Integration by Parts Trig Substitutions