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Antiderivative quicky

  • Thread starter Sympathy
  • Start date
10
0
1. Homework Statement
whats the integral of cos^2(x)



2. Homework Equations
?



3. The Attempt at a Solution

is it just sin^3(x)/3?
 

Answers and Replies

cepheid
Staff Emeritus
Science Advisor
Gold Member
5,183
35
No. And you know that that is the wrong answer because:

[tex] \frac{d}{dx}\left(\frac{\sin^3(x)}{3}\right) = \sin^2(x)\frac{d}{dx}(\sin x) = \sin^2(x)\cos(x) \not= \cos^2(x) [/tex]

Notice how I had a composition of functions and used the chain rule to differentiate. How do you deal with integrating something that has a composition of functions?
 
mjsd
Homework Helper
725
3
"easiest" way to integral this would probably be using the double angle formula
 
Gib Z
Homework Helper
3,344
4
Write out cos^2 x as (1-cos 2x)/2 is what mjsd meant.
 
integrtion by parts:

http://www.artofproblemsolving.com/Forum/latexrender/pictures/108554a051d314e3db05f4254cd279b0.gif [Broken]
 
Last edited by a moderator:
Gib Z
Homework Helper
3,344
4
That seems abit longer and harder than msjds method.
His way easily goes to this:
[tex]\frac{1}{2} \int 1 dx + \int \cos 2x dx = \frac{1}{2} (x + \int cos 2x dx)[/tex]

u=2x du/dx = 2
[tex]\frac{1}{2}(x+\frac{1}{2}\int cos u du)= \frac{1}{2} ( x+\frac{1}{2}\sin u)[/tex]


We're done, pretty fast too :)
[tex]\int \cos^2 x dx = \frac{x}{2} + \frac{\sin 2x}{4}[/tex]

BTW emilgouliev, nice work, and Welcome to Physicsforums. :)
 

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