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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?

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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
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$2\cos^2 x-1=\cos(2x)$, 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
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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
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1 Answer

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.

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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
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