Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Integrate $(\cos x)^4$. I see solutions using power reduction everywhere. I vaguely remember doing it based on some manipulation of trig identities $(\cos x)^2 = 1 - (\sin x)^2$ and $u$-substitution alone.

Anybody know what I am talking about?

share|cite|improve this question
The easiest way to do this, I think, is by writing $\cos(x)$ as $\frac{e^{ix}+e^{-ix}}{2}$ and expanding via the binomial theorem. – Brett Frankel Apr 28 '12 at 20:30
Or try to write $(\cos x)^4$ as a combination of $\cos 4x$, $\cos 3x$, $\cos 2x$ and $\cos x$, which turns out to be the same as what Brett just suggested :) – Thomas Andrews Apr 28 '12 at 20:31
$2\cos^2 x-1=\cos(2x)$, so we want to integrate $(1/4)(\cos(2x)+1)^2$. Expand, play similar but easier game with $\cos^2(2x)$ – André Nicolas Apr 28 '12 at 20:35
Just for future reference, this is often written as $\cos^4 x$ just to reduce the noise of parentheses. – Thomas Andrews Apr 28 '12 at 20:36
up vote 6 down vote accepted

Using basic trigonometric identities, we have

$$\begin{aligned}\cos^4(x)&=\cos^2(x)(1-\sin^2(x))\\ &=\cos^2(x)-\sin^2(x)\cos^2(x)\\ &=\cos^2(x)-\dfrac{\sin^2(2x)}{4}\\ &=\dfrac{1+\cos(2x)}{2}-\dfrac{1-\cos(4x)}{8},\end{aligned}$$

which should be much more manageable.

share|cite|improve this answer
Alternatively: $$\begin{align*}\cos^4 x&=\left(\frac{1+\cos\,2x}{2}\right)^2\\&=\frac14+\frac{\cos\,2x}{2}+\frac{\cos^2 2x}{4}\\&=\frac14+\frac{\cos\,2x}{2}+\frac14\frac{1+\cos\,4x}{2}\end{align*}$$ – J. M. Apr 29 '12 at 7:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.