JoeCMath

Sine and Cosine Power Integrals

Simple overview of integrals that are of the form \(\int \sin^n(x)\cos^m(x)dx\).

Main Idea

Consider the possible integrals of the following form

\[\int \sin^n(x)\cos^m(x)dx\]

where $n$ and $m$ are whole numbers greater than or equal to $1$.

There are four cases you can follow to calculate the integral that follow.

Watch this section: four cases at 0:12.

Case 1: Odd Sine Power

Consider the following

\[\int \sin^5(x)\cos^2(x)dx\]

where sine power is odd.

First put one copy of $\sin(x)$ to the side

\[\sin^5(x)=\sin^4(x)\sin(x)\]

Then the remaining even power of sine can be rewritten using

\[\sin^2(x)=1-\cos^2(x).\]

That gives

\[\sin^4(x)=\left(\sin^2(x)\right)^2 =\left(1-\cos^2(x)\right)^2\]

So the integral becomes

\[\int \left(1-\cos^2(x)\right)^2\cos^2(x)\sin(x)dx\]

Now we have $sin(x)$ off to the side, which will help us do u-substitution.

Let

\[u=\cos(x)\]

so

\[du=-\sin(x)dx\]

and

\[\sin(x)dx=-du\]

After substituting, the video gets

\[-\int (1-u^2)^2u^2 du\]

Expand the integral to use the power rule

\[(1-u^2)^2u^2=u^2-2u^4+u^6\]

Then integrate term by term

\[-\int \left(u^2-2u^4+u^6\right)\,du =-\left(\frac{u^3}{3}-\frac{2u^5}{5}+\frac{u^7}{7}\right)+C\]

Substitute $u=\cos(x)$ back in

\[-\frac{\cos^3(x)}{3} +\frac{2\cos^5(x)}{5} -\frac{\cos^7(x)}{7}+C\]

Watch this section: sine odd, cosine even at 0:28.

Case 2: Odd Cosine Power

Now, what happens when cosine is odd?

\[\int \sin^4(x)\cos^3(x) dx\]

Since cosine has the odd power, we put one $\cos(x)$ to the side

\[\cos^3(x)=\cos^2(x)\cos(x)\]

Then it rewrites the remaining cosine power with

\[\cos^2(x)=1-\sin^2(x)\]

The integral becomes

\[\int \sin^4(x)\left(1-\sin^2(x)\right)\cos(x) dx\]

With one $cos(x)$ to the side we are set up for u-substitution.

Let

\[u=\sin(x)\]

so

\[du=\cos(x) dx\]

Substitute and simplify

\[\int u^4(1-u^2) du =\int (u^4-u^6) du\]

Integrating gives

\[\frac{u^5}{5}-\frac{u^7}{7}+C\]

Return to $x$

\[\frac{\sin^5(x)}{5}-\frac{\sin^7(x)}{7}+C\]

Watch this section: sine even, cosine odd at 3:26.

Case 3: Both Powers Odd

What do we do when both powers are odd?

\[\int \sin^3(x)\cos^3(x) dx\]

If both powers are odd, you can do either case 1 or case 2 to find the integral.

Let’s keep one $\cos(x)$ on the side for this one.

We can rewrite

\[\cos^3(x)=\cos^2(x)\cos(x)\]

Then it converts the remaining cosine square:

\[\cos^2(x)=1-\sin^2(x)\]

So the integral becomes

\[\int \sin^3(x)\left(1-\sin^2(x)\right)\cos(x) dx\]

Let

\[u=\sin(x), \qquad du=\cos(x) dx\]

Then

\[\int u^3(1-u^2) du =\int (u^3-u^5) du\]

Integrate

\[\frac{u^4}{4}-\frac{u^6}{6}+C\]

Substitute back

\[\frac{\sin^4(x)}{4}-\frac{\sin^6(x)}{6}+C\]

Watch this section: both powers odd at 5:17.

Case 4: Both Powers Even

What happens when both powers are even?

\[\int \sin^2(x)\cos^2(x) dx\]

When both powers are even, there is no single $\sin(x)dx$ or $\cos(x)dx$ piece to save for an immediate u-substitution.

We switch to a power reduction method

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

and

\[\cos^2(x)=\frac{1+\cos(2x)}{2}\]

Substitute both

\[\int \sin^2(x)\cos^2(x) dx =\frac{1}{4}\int \left(1-\cos(2x)\right)\left(1+\cos(2x)\right) dx\]

Use the difference of squares

\[\left(1-\cos(2x)\right)\left(1+\cos(2x)\right) =1-\cos^2(2x)\]

So

\[\frac{1}{4}\int \left(1-\cos^2(2x)\right) dx\]

There is still an even power, so reduce again

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

After simplifying, we get

\[\frac{1}{8}\int \left(1-\cos(4x)\right) dx\]

Now integrate:

\[\frac{1}{8}\left(x-\frac{\sin(4x)}{4}\right)+C\]

The final answer is

\[\frac{x}{8}-\frac{\sin(4x)}{32}+C\]

Watch this section: both powers even at 6:44.

Video Summary

The video ends by returning to the same decision tree:

Case What to do
$n$ odd, $m$ even Save $\sin(x)$, convert sine powers to cosine, use $u=\cos(x)$.
$n$ even, $m$ odd Save $\cos(x)$, convert cosine powers to sine, use $u=\sin(x)$.
$n$ odd, $m$ odd Pick either odd-power method.
$n$ even, $m$ even Use power reduction identities.

The quick habit is: check the exponents first. Once you know which powers are odd or even, the next move is much less mysterious.

Watch this section: summary at 10:29.

Timestamp Guide

Section Main idea Video
Intro The video sets up $\int \sin^n(x)\cos^m(x)\,dx$. 0:00
Four cases The strategy depends on whether each power is odd or even. 0:12
Case 1 For $\int \sin^5(x)\cos^2(x)\,dx$, save $\sin(x)$ and use $u=\cos(x)$. 0:28
First final answer The result is $-\frac{\cos^3(x)}{3}+\frac{2\cos^5(x)}{5}-\frac{\cos^7(x)}{7}+C$. 2:52
Case 2 For $\int \sin^4(x)\cos^3(x)\,dx$, save $\cos(x)$ and use $u=\sin(x)$. 3:26
Case 3 When both powers are odd, either odd-power method can work. 5:17
Case 4 When both powers are even, use power reduction identities. 6:44
Even-even final answer $\int \sin^2(x)\cos^2(x)\,dx=\frac{x}{8}-\frac{\sin(4x)}{32}+C$. 10:03
Summary Check odd and even powers first, then choose the strategy. 10:29

This lesson uses u-substitution in the first three cases. For a slower review of that method, see the u-substitution topic page.

The examples are indefinite integrals, so every final answer includes $+C$. For that background, review the constant of integration notes.

The video description also points students toward the JoeCMath integration playlist: watch the integration playlist on YouTube.