Integrating cos^2(2x): Where Did I Go Wrong?

[Stratosphere]

Homework Statement

[tex]\int cos^{2}(2x) dx[/tex]

Homework Equations

The Attempt at a Solution

After doing all of the steps I came out with [tex]\frac{x+sin4x}{8}+c [/tex]
In the text it says the answer is [tex]\frac{x}{2}+\frac{sin4x}{8}+c [/tex]

Where did I go wrong?

 

[Cyosis]

While I and other people can guess, how do you expect us to point out the mistakes in your steps by only showing the answers? Show your steps.

 

[Stratosphere]

OK here are my steps then,
[tex]\cos^{2}=1/2(1+cos2x)[/tex]

[tex]\frac{1}{2}\int1+cos^{2}2x[/tex]

[tex]\frac{1}{8}(x+sin4x)[/tex]

 

[Cyosis]

Well the first error I can see is that your cosine squared doesn't have an argument at all which is obviously wrong. If you mean that the argument is x then your formula holds if the argument is 2x, as is shown in the integral, that formula does not hold.

Equation two is also wrong it says [itex]\cos^2 2x=1/2+1/2 \cos^22x[/itex], which again isn't true. You should use the first formula (the correct version of it) to replace the squared cosine so you end up without any squared trigonometric functions. Also don't forget the dx in your integration.

 

[Stratosphere]

I meant to write [tex]\ \frac{1}{2}\int cos(4x)[/tex] on the second one.

 

[Cyosis]

That is incorrect. If you use the first identity you've written down in its correct form you get [itex]\cos^2(2x)=\frac{1}{2}(1+\cos (4x))[/itex]. The left hand side is equal to the integrand so your integral becomes [itex]\frac{1}{2}\int 1+\cos (4x)dx [/itex]. Now can you show me the steps on how to integrate this?

 

[Stratosphere]

Integrate the one that then turns into an x and then the integral of cos(4x) is sin(4x), let u=4x, then du=4dx, after this is were I get confused, I think I'm suppose to take the 2 and divide everything by it which would make all of the integral times 1/8, but that must be wrong since I didn't get the right answer.

 

[Cyosis]

The first thing you need to teach yourself is being accurate. You've put wrong and sloppy info in each of your post so far. First off [itex]\int \cos(4x) dx \neq \sin (4x)[/itex]. The substitution you list however is the correct way of doing it. Sow me how you worked out the entire substitution and how dividing all the terms by 2 would cause the entire integral to be multiplied by 1/8. You're pretty close, but you're being very sloppy. I am quite sure that if you write it out neatly you will notice the mistake all by yourself.

 

[Stratosphere]

Oh, I see what I did wrong there, thanks.