# Evaluating $\int \cos^4(x)\operatorname d\!x$

I want to Evaluate integral :$$\int \cos^4(x)\operatorname d\!x$$ And think about doing the following thing: $$\int \left(1-\sin^2(x)\right)^2\operatorname d\!x \to \int \left(1-2\sin^2(x)+\sin^4(x)\right)\operatorname d\!x$$ but I think I just complicated it.

Any suggestions?
Thanks!

• – vadim123 May 8 '13 at 20:41
• Bigger hint: $\cos^4x = (\cos^2 x)^2=(\frac{1+\cos 2x}{2})^2$ – vadim123 May 8 '13 at 20:42

## 5 Answers

$$\cos^4(x) = \left(\dfrac{1+\cos(2x)}2 \right)^2 = \dfrac{1 + \cos^2(2x) + 2\cos(2x)}4 = \dfrac{1 + \dfrac{1+\cos(4x)}2 + 2\cos(2x)}4$$ which gives us $$\cos^4(x) = \dfrac{3 + 4 \cos(2x) + \cos(4x)}8$$ Now you should be able to integrate this off.

Using the reduction formulae,

$$\int\cos^nxdx=\frac{\cos^{n-1}x\sin x}n+\frac{n-1}n \int\cos^{n-2}xdx$$

Putting $n=2,$ $$\int\cos^2xdx=\frac{\cos x\sin x}2+\frac12 \int dx=\frac{\cos x\sin x}2+\frac12 x+C$$

Putting $n=4,$ $$\int\cos^4xdx=\frac{\cos^3x\sin x}4+\frac34 \int\cos^2xdx$$

$$\cos^4x=\cos^2x-\cos^2x\sin^2x\implies$$

$$\implies\text{I}:=\int\cos^4x\,dx=\int\cos^2xdx-\int\cos^2x\sin^2xdx=$$

$$\frac{x+\cos x\sin x}2+\int\sin x\cos^2x\,(-\cos)' xdx$$

Now by parts in the last integral::

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

so

$$\text{I}:=\frac{x\cos x\sin x}2-\frac13\cos^3x\sin x+\frac13\text{I}\implies\;\;\ldots$$

As indicated in the source pointed out by Vadim123 in the comments,

$\displaystyle \cos^4x=\frac{3 + 4 \cos2x + \cos4x}{8}$

Plug this in and integrate.

One more (using $\cos^4 x = \cos^2 x \cos^2 x = (1-\sin^2 x) \cos^2x$): $$I=\int \cos^4 x dx = \int \cos^2x dx - \int (\sin x \cos x)^2dx=\int \cos^2 x dx\\ -\frac{1}{4} \int (\sin2x)^2dx=\int \cos^2 x dx-\frac{1}{8}\int \cos^2tdt$$ then use the power reduction formulas: $$\sin^2 x =\frac{1-\cos2x}{2}\\ \cos^2 x=\frac{1+\cos 2x}{2}$$ Can you handle from here?