What is the integral of #cos^6(x)#?
2 Answers
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
#int (1+cos(2x))/2 * (1+cos(2x))/2 * (1+cos(2x))/2 dx #
You can bring the factor
#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:
Whenever you have a basic integral (like cos), but with a different
#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 = int (1 + cos(4x))/2#
(using the identity. It becomes
#= 1/2int dx + 1/2int cos(4x)dx#
#= 1/2x + 1/2sin(4x)*1/4#
#= 1/2x + 1/8sin(4x)#
Next,
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
#int (1-sin^2(2x))cos(2x)dx#
Now you should apply
#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:
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!
I would like to present an alternative approach - based on the idea that "complex" is simple!
Explanation:
We know that
and that
Using these we get
We now use standard integrals to evaluate :