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QUESTION

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

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)##

##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##

##int (1-u^2)du## ##int du - int u^2 du## ##u - 1/3u^3## ##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(sin(2x)-1/3sin^3(2x))##

##=1/8x + 3/16sin(2x) + 3/16x + 3/64sin(4x) + 1/8sin(2x)-1/24sin^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!