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Cos^2(x) Integral..

by iRaid
Tags: cos2x, integral
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iRaid
#1
Apr12-12, 10:49 PM
P: 550
1. The problem statement, all variables and given/known data
[itex]\int cos^{2}x dx[/itex]

I know that

[itex]cos^{2}x = \frac{1+cos2x}{2}[/itex]

but I don't see how that helps me.
Can someone help walk me through it..

2. Relevant equations



3. The attempt at a solution
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QuarkCharmer
#2
Apr12-12, 10:53 PM
QuarkCharmer's Avatar
P: 1,035
Well, that leaves you with:
[tex]\int \frac{1}{2}+\frac{1}{2}cos(2x)dx[/tex]

Which you can break up into two integrals:
[tex]\int \frac{1}{2}dx + \int \frac{1}{2}cos(2x)dx[/tex]

The first one should be no problem. Isn't there some sort of substitution you can make for the second one?
Dick
#3
Apr12-12, 10:53 PM
Sci Advisor
HW Helper
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P: 25,246
Can you integrate cos(2x)? Use a u substitution.

iRaid
#4
Apr12-12, 11:03 PM
P: 550
Cos^2(x) Integral..

u=2x du=(1/2)dx
(1/2)∫cosudu
=(1/4)sin2x

So then..
x/2 + (1/4)sin2x
but thats not the answer..
Dick
#5
Apr12-12, 11:05 PM
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HW Helper
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P: 25,246
Quote Quote by iRaid View Post
u=2x du=(1/2)dx
(1/2)∫cosudu
=(1/4)sin2x

So then..
x/2 + (1/4)sin2x
but thats not the answer..
I think it is the correct answer. You should probably put a +C on it. Is that the problem?
iRaid
#6
Apr12-12, 11:06 PM
P: 550
Quote Quote by Dick View Post
I think it is the correct answer. You should probably put a +C on it. Is that the problem?
Oh nevermind was looking at the wrong answer. Thanks for the help.


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