Solving Trig Integrals help

[Totalderiv]

Homework Statement

$\int cos^6(x)$

Homework Equations

$1 = sin^2(x) + cos^2(x)$

The Attempt at a Solution

$\int cos(x) * cos^5(x)$
$\int cos(x) * (cos^2(x))^3$
$\int cos(x) * (1-sin^2(x))^3$

This is where I got lost, we just started this topic and I have a lot of homework to do, but I can't get past this question. Please help!

 

[lineintegral1]

Your second step is not algebraically correct. Check your exponential rules. You are using the right strategy for solving this integral, but is there another breakdown of powers you can try instead of first and fifth powers of cosine?

 

[Totalderiv]

I could either do

$\int cos^3(x) * cos^3(x)$

or

$\int (cos^2(x))^2 * cos^2(x)$

or

$\int cos^4(x) * cos^2(x)$

I have tried all of these, but couldn't get an answer unless I'm supposed to use the half-angle identities. Any one have a clue on how I should split this up?

 

[dynamicsolo]

I could either do

$\int cos^3(x) * cos^3(x)$

or

$\int (cos^2(x))^2 * cos^2(x)$

or

$\int cos^4(x) * cos^2(x)$

I have tried all of these, but couldn't get an answer unless I'm supposed to use the half-angle identities. Any one have a clue on how I should split this up?

You won't be able to get any of those to work, since you will have difficulty locating something to use as a differential for the integral.

When you have even powers of sine or cosine, we use a handy "trick"* based on the "double-angle formula" for cosine:

$$\cos(2 \theta) = \cos^{2}\theta - \sin^{2}\theta = 1 - 2\sin^{2}\theta = 2\cos^{2}\theta - 1 ,$$

these last two forms coming from applying the Pythagorean Identity. We now have

$$\cos(2 \theta) = 1 - 2\sin^{2}\theta \Rightarrow \sin^{2}\theta = \frac{1}{2} [ 1 - \cos(2 \theta) ]$$

and

$$\cos(2 \theta) = 2\cos^{2}\theta - 1 \Rightarrow \cos^{2}\theta = \frac{1}{2} [ 1 + \cos(2 \theta) ]$$

*A "trick" is a method that only works for certain special problems...

You now write your integral as $\int cos^{6} x dx = \int ( cos^{2} x )^{3} dx$ , and substitute the expression for "cosine-squared". You are going to get a polynomial with powers of cos(2x) ; for the term [cos(2x)]² , you use this substitution again , which is going to lead to a term with cos(2 · 2x) or cos(4x) , and so on.

As a hint of what to expect, even powers of sine or cosine lead to a sum of terms involving sine or cosine of even multiples of the original angle. You will have terms up to cos(6x) here. But the point of using this "trick" is that we do know how to integrate cos(kx)...