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 #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!

Apr 22, 2018

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)#