
#1
Apr1212, 10:49 PM

P: 542

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 



#2
Apr1212, 10:53 PM

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? 



#3
Apr1212, 10:53 PM

Sci Advisor
HW Helper
Thanks
P: 25,171

Can you integrate cos(2x)? Use a u substitution.




#4
Apr1212, 11:03 PM

P: 542

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.. 



#5
Apr1212, 11:05 PM

Sci Advisor
HW Helper
Thanks
P: 25,171





#6
Apr1212, 11:06 PM

P: 542




Register to reply 
Related Discussions  
Does path integral and loop integral in a Feynman diagram violate special relativity?  General Physics  0  
Integral equation with a derivative of the function inside the integral  Calculus & Beyond Homework  5  
Rewrite the integral as an equivalent iterated integral in the order  Calculus & Beyond Homework  5  
Using polar coord. to change double integral into single integral involving only r.  Calculus & Beyond Homework  5  
Is the ordinary integral a special case of the line integral?  Calculus  3 