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Here my attempts for integrating $\cos^4(x)$ in few methods.

1st method.

$(\cos^2x)^2=(\frac{1}{2})^2(1+\cos2x)^2$

$=\frac{1}{4}(1+2\cos2x+\cos^22x)=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\cos^22x)$

$=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\frac{1}{2}(1+\cos4x))$

$=\frac{3}{8}+\frac{1}{2}\cos2x+\frac{1}{8}\cos4x$

So, $\int\cos^4xdx=\int(\cos^2x)^2dx$

$=\int\frac{3}{8}dx+\frac{1}{2}\int\cos2xdx+\frac{1}{8}\int\cos4xdx$

$=\frac{1}{32}(12x+8\sin2x+\sin4x)+c$

2nd method.

By using reduction formula, $\cos^4xdx=\frac{1}{4}\sin x\cos^3x+\frac{3}{4}\int\cos^2xdx$

$=\frac{1}{4}\sin x\cos^3x+\frac{3}{4}\int(\frac{1}{2}\cos2x+\frac{1}{2})dx$

$=\frac{1}{4}\sin x\cos^3x+\frac{3}{8}\int\cos2x+\frac{3}{8}\int1dx$

$=\frac{1}{4}\sin x\cos^3x+\frac{3}{16}\sin 2x+\frac{3}{8}x$

$=\frac{1}{4}\sin x \cos^3x+\frac{3}{8}\sin x \cos x+\frac{3}{8}x$

3rd method, By using De Moivre's formula.

4th method

$\int \cos^4xdx=\int (\frac{e^{ix}+e^{-ix}}{2})^4dx$

$=\frac{1}{16}\int (e^{4ix}+4e^{2ix}+6+4e^{-2ix}+e^{-4ix})dx$

$=\frac{1}{16}\int (e^{4ix}+e^{-4ix}+4(e^{2ix}+e^{-2ix})+6)dx$

$=\frac{1}{16}(\frac{1}{4i}(e^{4ix}-e^{-4ix})+\frac{2}{i}(e^{2ix}-e^{-2ix})+6x)+c$

$\frac{1}{16}(\frac{1}{2}(\frac{e^{4ix}-e^{-4ix}}{2i})+4(\frac{e^{2ix}-e^{-2ix}}{2i})+6x)+c$

$\frac{1}{16}(\frac{1}{2}\sin 4x+4\sin 2x+6x)+c$

$\frac{1}{8}(\frac{1}{4}\sin 4x+2\sin 2x+3x)+c$

I want to ask is there more method to integrate it? Thanks.

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  • $\begingroup$ If you know how to integrate $ \sin^{2} $, then you’re done because \begin{align} {\cos^{4}}(x) & = {\cos^{2}}(x) \cdot \left[ 1 - {\sin^{2}}(x) \right] \\ & = {\cos^{2}}(x) - {\cos^{2}}(x) \cdot {\sin^{2}}(x) \\ & = {\cos^{2}}(x) - \frac{1}{4} [2 \sin(x) \cos(x)]^{2} \\ & = \left[ 1 - {\sin^{2}}(x) \right] - \frac{1}{4} {\sin^{2}}(2 x). \end{align} $\endgroup$ – Berrick Caleb Fillmore Apr 23 '15 at 7:51
  • $\begingroup$ I think you've covered all the obvious methods unless you want to express it as a power series and integrate term by term...................... $\endgroup$ – Mathemagician1234 Apr 23 '15 at 7:56
  • $\begingroup$ @Mathemagician1234: The OP wants to hack this problem into pieces. $\endgroup$ – Berrick Caleb Fillmore Apr 23 '15 at 7:59
  • $\begingroup$ $cos^2(x)=\frac {1+cos(2x)}2$ and $sin^2(x)=\frac {1-cos(2x)}2$ Then $cos^4(x)=\frac {1+cos(2x)}2 - \frac 14 \frac {1-cos(4x)}2$ $\endgroup$ – user128766 Apr 23 '15 at 8:06
  • $\begingroup$ @Mathemagician1234 I don't know why. Whenever I solving problem, I will have to know methodss to solve it. $\endgroup$ – Mathxx Apr 23 '15 at 8:18
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Let us first establish a general formula, with an obvious notation:

$$(C^cS^s)'=-cC^{c-1}S^{s+1}+sC^{c+1}S^{s-1}=-cC^{c-1}(1-C^2)S^{s-1}+sC^{c+1}S^{s-1}\\ =-cC^{c-1}S^{s-1}+(s+c)C^{c+1}S^{s-1}.$$

Then, rewriting and shifting the indexes,

$$\int C^\color{green}cS^s\,dx=\frac1{s+c}C^{c-1}S^{s+1}+\frac{c-1}{s+c}\int C^{\color{green}{c-2}}S^s\,dx.$$

Then $$C^2\to\frac12CS+\frac12x,$$ $$C^4\to\frac14C^3S+\frac34\left(\frac12CS+\frac12x\right).$$

Deriving $C^{c-1}S$ directly leads to the recurrence $$\int C^\color{green}c\,dx=\frac1cC^{c-1}S+\frac{c-1}c\int C^{\color{green}{c-2}}\,dx.$$

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I think you gave the most straightforward methods, including the sure-to-success replacement by $e^{ix}+e^{-ix}$. Basically any other method is going to be a more obfuscated form of this one.

You could also give a shot to the rational parametrization of the unit circle, by using $\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2\,dt}{1+t^2}$, so that $\int \cos^4 x dx = \int \frac {2(1-t^2)^4dt}{(1+t^2)^5}$, expanding this ugly rational fraction at the poles, and then reusing the rational parametrization to fall back on the trigonometric result.

There might also be a way to do this using Fourier expansion, but I'm not sure I want to see this.

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