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$$\int \cos^4 x\,\mathrm dx = \int (1 - \sin^2x)^2\,\mathrm dx.$$

I tried using $\cos^2(x) + \sin^2(x)$ = 1. This was on the integration by parts section of my textbook.

The integral I came out with is given me a hard time. I double checked my work and I feel I am on the right track but I feel stuck. I tried u substitution but there is no $\cos x$ to leverage with.

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This is really the effect of what AWertheim is saying, and not that difficult. $$ \begin{align*} \cos^4x &= \left(\cos^2x\right)^2 \\ &=\left( \frac{1+\cos(2x)}{2} \right)^2 \\ &=\frac{1}{4}\left( 1+2\cos(2x)+\cos^2(2x) \right) \\ &= \frac{1}{4}\left( 1+2\cos(2x)+\frac{1+\cos(4x)}{2} \right) \\ &=\frac{1}{4}+\frac{\cos(2x)}{2}+\frac{1}{8}+\frac{\cos(4x)}{8} \\ &=\frac{3}{8}+\frac{\cos(2x)}{2}+\frac{\cos(4x)}{8}. \end{align*} $$

The sole advantage here being that this form is very simple to integrate.

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    $\begingroup$ Very clear exposition! +1 :) $\endgroup$ – Alex Wertheim Oct 5 '14 at 4:28
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Hint: apply the half angle formula

$$\cos^{2}(x) = \frac{1}{2}\left(1+\cos(2x)\right)$$

twice.

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$$\int\cos^4xdx=\int (1-\sin^2x)\cos^2x=\int\cos^2xdx-\int\cos^2x\sin^2xdx$$

$$\begin{align*}(1)&\;\int\cos^2xdx=\frac{x+\sin x\cos x}2+C\\{}\\ (2)&\;\int\cos^2x\sin^2x dx=\int\cos^2x\sin x\cdot\sin xdx\end{align*}$$

and now by parts:

$$u=\sin x\;,\;\;u'=\cos x\\ v'=\cos^2x\sin x\;,\;\;v=-\frac13\cos^3x$$

Try to take it from here.

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Hint

You can also use power reduction formulas $$\cos^2(x)=\frac{1+\cos(2x)}{2}$$ $$\cos^3(x)=\frac{3\cos(x)+\cos(3x)}{4}$$ $$\cos^4(x)=\frac{3+4\cos(2x)+\cos(4x)}{8}$$ $$\cos^5(x)=\frac{10\cos(x)+5\cos(3x)+\cos(5x)}{16}$$

There are similar formulas for $\sin^n(x)$, $\sinh^n(x)$, $\cosh^n(x)$

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In general$\newcommand{\dx}{\,\mathbb d x}$\begin{align*} I_n = \int \cos^n x \dx &= \int (1-\sin^2 x)\cos^{n-2} x \dx \\ &= I_{n-2} - \int \underbrace{\sin x \cos^{n-2}x}_{v^\prime} \underbrace{\sin x}_u \dx \\ &= I_{n-2} + \frac{1}{n-1}\cos^{n-1} x \sin x - \frac{1}{n-1}\int \cos^{n-1} x \cos x \dx \\ &= I_{n-2} + \frac{1}{n-1}\cos^{n-1} x \sin x - \frac{1}{n-1}I_n \\ \frac{n}{n-1} I_n &= I_{n-2} + \frac{1}{n-1}\cos^{n-1} x \sin x \\[12pt] I_n &= \frac{n-1}{n}I_{n-2} + \frac{1}{n}\cos^{n-1} x \sin x \end{align*} Valid for $n \gt 1$

So \begin{align*}\int \cos^4 x \dx &= \frac34\int\cos^2 x \dx + \frac13\cos^3 x \sin x \\ &= \frac34\left( \frac12\int\dx+\frac12\cos x \sin x \right) + \frac13\cos^3 x \sin x \\ &= \frac38x + \frac38 \cos x \sin x + \frac13 \cos^3 x \sin x +c\end{align*}

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