Question

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

Answer 1

See explanation.

Explanation:

This will be a long answer.

So what you want to find is:

`int cos^6(x)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(x) = (1+cos(2x))/2`

First we split up the cosines:

`int cos^2(x)*cos^2(x)*cos^2(x) dx`

Now we can replace every `cos^2(x)` with the identity above:

`int (1+cos(2x))/2 * (1+cos(2x))/2 * (1+cos(2x))/2 dx`

You can bring the factor `1/8` out of the integral:

`1/8 int (1+cos(2x)) * (1+cos(2x)) * (1+cos(2x)) dx`

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

`(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3`

Let's apply this to the integral.

`1/8 int (1+cos(2x))^3dx`

`=1/8 int 1^3+3*1^2*cos(2x)+3*1*cos^2(2x)+cos^3(2x) dx`

`=1/8 int 1+3cos(2x)+3cos^2(2x)+cos^3(2x) dx`

Now we can already splice this integral up a bit:

`1/8(int 1dx + 3int cos(2x)dx + 3int cos^2(2x)dx + intcos^3(2x)dx)`

`1/8x+ 3/16sin(2x) + 1/8(3int cos^2(2x)dx + intcos^3(2x)dx)`

If you need to know how I instantly came up with that second term:
`int cos(2x)dx`

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

`sin(2x)*1/2`

Back to the problem: we will remember the first two factors of the solution and we will solve `int cos^2(2x)dx` and `int cos^3(2x)dx` seperately.

`int cos^2(2x)dx = int (1 + cos(4x))/2`

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

`= 1/2int dx + 1/2int cos(4x)dx`

`= 1/2x + 1/2sin(4x)*1/4`

`= 1/2x + 1/8sin(4x)`

Next, `int cos^3(2x)dx`

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

`int cos^2(2x)cos(2x)dx`

Now you should use the identity `sin^2(x)+cos^2(x) = 1`

`int (1-sin^2(2x))cos(2x)dx`

Now you should apply `u`-substitution:

`u = sin(2x) <=> du = 2cos(2x)dx <=> 1/2 du = cos(2x)dx`

So

`1/2int (1-u^2)du`

`1/2int du - int u^2 du`

`1/2(u - 1/3u^3)`

`1/2[sin(2x)-1/3sin^3(2x)]`

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

`1/8x+ 3/16sin(2x) + 1/8(3int cos^2(2x)dx + intcos^3(2x)dx)`
`= 1/8x + 3/16sin(2x) + 3/8[(1/2x + 1/8sin(4x)) + 1/8[1/2 * (sin(2x)-1/3sin^3(2x))]`

`=1/8x + 3/16sin(2x) + 3/16x + 3/64sin(4x) + 1/16sin(2x)-1/48sin^3(2x)`

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!

Answer 2

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

Explanation:

We know that

`cos(x) = (e^{ix}+e^{-ix})/2`

and that

`(a+b)^6 = a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6`

Using these we get

`cos^6(x) = ((e^{ix}+e^{-ix})/2)^6`
`qquad = 1/2^6[(e^{ix})^6+6(e^{ix})^5e^{-ix}+15(e^{ix})^4(e^{-ix})^2`
`qquad qquad+ 20(e^{ix})^3(e^{-ix})^3 +15(e^{ix})^2(e^{-ix})^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}]`
`qquad = 1/2^6[(e^{i6x}+e^{-i6x})+6(e^{i4x}+e^{-i4x})+15(e^{i2x}+e^{-i2x})+20]`
`qquad = 1/32 cos(6x)+3/16 cos(4x)+15/32 cos(2x)+5/16`

We now use standard integrals to evaluate :

`int cos^6(x)dx = int [1/32 cos(6x)+3/16 cos(4x)+15/32 cos(2x)+5/16]dx`
`qquad = color(red)(1/192 sin(6x)+3/64 sin(4x)+15/64sin(2x)+5/16x+C)`