- #1

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## Homework Statement

whats the integral of cos^2(x)

## Homework Equations

?

## The Attempt at a Solution

is it just sin^3(x)/3?

- Thread starter Sympathy
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- #1

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whats the integral of cos^2(x)

?

is it just sin^3(x)/3?

- #2

cepheid

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[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?

- #3

mjsd

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"easiest" way to integral this would probably be using the double angle formula

- #4

Gib Z

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Write out cos^2 x as (1-cos 2x)/2 is what mjsd meant.

- #5

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integrtion by parts:

http://www.artofproblemsolving.com/Forum/latexrender/pictures/108554a051d314e3db05f4254cd279b0.gif [Broken]

http://www.artofproblemsolving.com/Forum/latexrender/pictures/108554a051d314e3db05f4254cd279b0.gif [Broken]

Last edited by a moderator:

- #6

Gib Z

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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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