# What is the integral of cos^6(x)?

Dec 19, 2014

See explanation.

#### Explanation:

This will be a long answer.

So what you want to find is:

$\int {\cos}^{6} \left(x\right) \mathrm{dx}$

There's a rule of thumb that you can remember: whenever you need to integrate an even power of the cosine function, you need to use the identity:

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

First we split up the cosines:

$\int {\cos}^{2} \left(x\right) \cdot {\cos}^{2} \left(x\right) \cdot {\cos}^{2} \left(x\right) \mathrm{dx}$

Now we can replace every ${\cos}^{2} \left(x\right)$ with the identity above:

$\int \frac{1 + \cos \left(2 x\right)}{2} \cdot \frac{1 + \cos \left(2 x\right)}{2} \cdot \frac{1 + \cos \left(2 x\right)}{2} \mathrm{dx}$

You can bring the factor $\frac{1}{8}$ out of the integral:

$\frac{1}{8} \int \left(1 + \cos \left(2 x\right)\right) \cdot \left(1 + \cos \left(2 x\right)\right) \cdot \left(1 + \cos \left(2 x\right)\right) \mathrm{dx}$

Now you could apply FOIL twice, but I would rather use Newton's Binomial theorem. Following from this theorem is that

${\left(x + y\right)}^{3} = {x}^{3} + 3 {x}^{2} y + 3 x {y}^{2} + {y}^{3}$

Let's apply this to the integral.

$\frac{1}{8} \int {\left(1 + \cos \left(2 x\right)\right)}^{3} \mathrm{dx}$

$= \frac{1}{8} \int {1}^{3} + 3 \cdot {1}^{2} \cdot \cos \left(2 x\right) + 3 \cdot 1 \cdot {\cos}^{2} \left(2 x\right) + {\cos}^{3} \left(2 x\right) \mathrm{dx}$

$= \frac{1}{8} \int 1 + 3 \cos \left(2 x\right) + 3 {\cos}^{2} \left(2 x\right) + {\cos}^{3} \left(2 x\right) \mathrm{dx}$

Now we can already splice this integral up a bit:

$\frac{1}{8} \left(\int 1 \mathrm{dx} + 3 \int \cos \left(2 x\right) \mathrm{dx} + 3 \int {\cos}^{2} \left(2 x\right) \mathrm{dx} + \int {\cos}^{3} \left(2 x\right) \mathrm{dx}\right)$

$\frac{1}{8} x + \frac{3}{16} \sin \left(2 x\right) + \frac{1}{8} \left(3 \int {\cos}^{2} \left(2 x\right) \mathrm{dx} + \int {\cos}^{3} \left(2 x\right) \mathrm{dx}\right)$

If you need to know how I instantly came up with that second term:
$\int \cos \left(2 x\right) \mathrm{dx}$

Whenever you have a basic integral (like cos), but with a different $x$ ($a x$), you can just integrate normally, but in the end, multiply by a factor of $\frac{1}{a}$. Here it becomes:

$\sin \left(2 x\right) \cdot \frac{1}{2}$

Back to the problem: we will remember the first two factors of the solution and we will solve $\int {\cos}^{2} \left(2 x\right) \mathrm{dx}$ and $\int {\cos}^{3} \left(2 x\right) \mathrm{dx}$ seperately.

$\int {\cos}^{2} \left(2 x\right) \mathrm{dx} = \int \frac{1 + \cos \left(4 x\right)}{2}$

(using the identity. It becomes $4 x$ because you double it.)

$= \frac{1}{2} \int \mathrm{dx} + \frac{1}{2} \int \cos \left(4 x\right) \mathrm{dx}$

$= \frac{1}{2} x + \frac{1}{2} \sin \left(4 x\right) \cdot \frac{1}{4}$

$= \frac{1}{2} x + \frac{1}{8} \sin \left(4 x\right)$

Next, $\int {\cos}^{3} \left(2 x\right) \mathrm{dx}$

Whenever you have an odd power of cosines, you can do the following:

$\int {\cos}^{2} \left(2 x\right) \cos \left(2 x\right) \mathrm{dx}$

Now you should use the identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

$\int \left(1 - {\sin}^{2} \left(2 x\right)\right) \cos \left(2 x\right) \mathrm{dx}$

Now you should apply $u$-substitution:

$u = \sin \left(2 x\right) \iff \mathrm{du} = 2 \cos \left(2 x\right) \mathrm{dx} \iff \frac{1}{2} \mathrm{du} = \cos \left(2 x\right) \mathrm{dx}$

So

$\frac{1}{2} \int \left(1 - {u}^{2}\right) \mathrm{du}$

$\frac{1}{2} \int \mathrm{du} - \int {u}^{2} \mathrm{du}$

$\frac{1}{2} \left(u - \frac{1}{3} {u}^{3}\right)$

$\frac{1}{2} \left[\sin \left(2 x\right) - \frac{1}{3} {\sin}^{3} \left(2 x\right)\right]$

Now we have all our parts to complete the integral. Remember that we had:

$\frac{1}{8} x + \frac{3}{16} \sin \left(2 x\right) + \frac{1}{8} \left(3 \int {\cos}^{2} \left(2 x\right) \mathrm{dx} + \int {\cos}^{3} \left(2 x\right) \mathrm{dx}\right)$
= 1/8x + 3/16sin(2x) + 3/8[(1/2x + 1/8sin(4x)) + 1/8[1/2 * (sin(2x)-1/3sin^3(2x))]

$= \frac{1}{8} x + \frac{3}{16} \sin \left(2 x\right) + \frac{3}{16} x + \frac{3}{64} \sin \left(4 x\right) + \frac{1}{16} \sin \left(2 x\right) - \frac{1}{48} {\sin}^{3} \left(2 x\right)$

You could simplify this a bit, which isn't that hard, I'll leave that as a challenge to you :D.

I hope this helps. It was fun!

Apr 22, 2018

I would like to present an alternative approach - based on the idea that "complex" is simple!

#### Explanation:

We know that

$\cos \left(x\right) = \frac{{e}^{i x} + {e}^{- i x}}{2}$

and that

${\left(a + b\right)}^{6} = {a}^{6} + 6 {a}^{5} b + 15 {a}^{4} {b}^{2} + 20 {a}^{3} {b}^{3} + 15 {a}^{2} {b}^{4} + 6 a {b}^{5} + {b}^{6}$

Using these we get

${\cos}^{6} \left(x\right) = {\left(\frac{{e}^{i x} + {e}^{- i x}}{2}\right)}^{6}$
qquad = 1/2^6[(e^{ix})^6+6(e^{ix})^5e^{-ix}+15(e^{ix})^4(e^{-ix})^2
$q \quad q \quad + 20 {\left({e}^{i x}\right)}^{3} {\left({e}^{- i x}\right)}^{3} + 15 {\left({e}^{i x}\right)}^{2} {\left({e}^{- i x}\right)}^{4}$
qquad qquad +6(e^{ix})^5e^{-ix}+(e^{-ix})^6 ]
qquad = 1/2^6[e^{i6x}+6e^{i4x}+15e^{i2x}+20
qquad qquad +15e^{-i2x}+6e^{-i4x}+e^{-i6x}]
$q \quad = \frac{1}{2} ^ 6 \left[\left({e}^{i 6 x} + {e}^{- i 6 x}\right) + 6 \left({e}^{i 4 x} + {e}^{- i 4 x}\right) + 15 \left({e}^{i 2 x} + {e}^{- i 2 x}\right) + 20\right]$
$q \quad = \frac{1}{32} \cos \left(6 x\right) + \frac{3}{16} \cos \left(4 x\right) + \frac{15}{32} \cos \left(2 x\right) + \frac{5}{16}$

We now use standard integrals to evaluate :

$\int {\cos}^{6} \left(x\right) \mathrm{dx} = \int \left[\frac{1}{32} \cos \left(6 x\right) + \frac{3}{16} \cos \left(4 x\right) + \frac{15}{32} \cos \left(2 x\right) + \frac{5}{16}\right] \mathrm{dx}$
$q \quad = \textcolor{red}{\frac{1}{192} \sin \left(6 x\right) + \frac{3}{64} \sin \left(4 x\right) + \frac{15}{64} \sin \left(2 x\right) + \frac{5}{16} x + C}$