Sine and Cosine Power Integrals
This page follows the JoeCMath video on integrating powers of sine and cosine. The video is about choosing a strategy before doing algebra: check whether the powers are odd or even, then decide whether to save a trig factor for u-substitution or use power reduction.
Main Idea
The video studies integrals of the form
\[\int \sin^n(x)\cos^m(x)\,dx,\]where $n$ and $m$ are whole numbers at least $1$.
The first decision is not to expand everything. The first decision is to inspect the powers. If the sine power is odd, the video saves one $\sin(x)$, converts the remaining sine powers to cosines, and uses $u=\cos(x)$. If the cosine power is odd, the video saves one $\cos(x)$, converts the remaining cosine powers to sines, and uses $u=\sin(x)$.
When both powers are odd, either of those odd-power strategies can work. When both powers are even, the video switches away from that u-substitution setup and uses power reduction identities instead.
Watch this section: four cases at 0:12.
Case 1: Odd Sine Power
The first example in the video is
\[\int \sin^5(x)\cos^2(x)\,dx.\]The sine power is odd, so the video saves one copy of $\sin(x)$:
\[\sin^5(x)=\sin^4(x)\sin(x).\]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 the substitution is set up:
\[u=\cos(x),\]so
\[du=-\sin(x)\,dx\]and
\[\sin(x)\,dx=-du.\]After substituting, the video gets
\[-\int (1-u^2)^2u^2\,du.\]Expand first:
\[(1-u^2)^2u^2=u^2-2u^4+u^6.\]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:
\[-\frac{\cos^3(x)}{3} +\frac{2\cos^5(x)}{5} -\frac{\cos^7(x)}{7}+C.\]The key move is saving the single $\sin(x)\,dx$ piece so it can pair with $du=-\sin(x)\,dx$.
Watch this section: sine odd, cosine even at 0:28.
Case 2: Odd Cosine Power
The second example flips the odd power:
\[\int \sin^4(x)\cos^3(x)\,dx.\]Since cosine has the odd power, the video saves one $\cos(x)$:
\[\cos^3(x)=\cos^2(x)\cos(x).\]Then it rewrites the remaining cosine power with
\[\cos^2(x)=1-\sin^2(x).\]The integral becomes
\[\int \sin^4(x)\left(1-\sin^2(x)\right)\cos(x)\,dx.\]This time the substitution is
\[u=\sin(x),\]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 $x$:
\[\frac{\sin^5(x)}{5}-\frac{\sin^7(x)}{7}+C.\]The pattern mirrors the first case: when cosine has the odd power, save one $\cos(x)$ and convert the rest into sine.
Watch this section: sine even, cosine odd at 3:26.
Case 3: Both Powers Odd
The third example is
\[\int \sin^3(x)\cos^3(x)\,dx.\]Because both powers are odd, the video points out that either previous method can work. In the worked version, it saves one $\cos(x)$:
\[\cos^3(x)=\cos^2(x)\cos(x).\]Then it converts the remaining cosine square:
\[\cos^2(x)=1-\sin^2(x).\]So the integral becomes
\[\int \sin^3(x)\left(1-\sin^2(x)\right)\cos(x)\,dx.\]Use
\[u=\sin(x), \qquad du=\cos(x)\,dx.\]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:
\[\frac{\sin^4(x)}{4}-\frac{\sin^6(x)}{6}+C.\]The useful takeaway is that both-odd cases give you a choice. Save either one sine factor or one cosine factor, then convert the remaining even power with a Pythagorean identity.
Watch this section: both powers odd at 5:17.
Case 4: Both Powers Even
The fourth example is different:
\[\int \sin^2(x)\cos^2(x)\,dx.\]When both powers are even, there is no single $\sin(x)\,dx$ or $\cos(x)\,dx$ piece to save for an immediate u-substitution. The video switches to power reduction:
\[\sin^2(x)=\frac{1-\cos(2x)}{2}\]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, the video reaches
\[\frac{1}{8}\int \left(1-\cos(4x)\right)\,dx.\]Now integrate:
\[\frac{1}{8}\left(x-\frac{\sin(4x)}{4}\right)+C.\]The final answer is
\[\frac{x}{8}-\frac{\sin(4x)}{32}+C.\]The main difference in the even-even case is that power reduction does the work that u-substitution did in the odd-power cases.
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
Related Calculus Work
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.