# Antiderivative quicky

Sympathy

## Homework Statement

whats the integral of cos^2(x)

?

## The Attempt at a Solution

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

Staff Emeritus
Gold Member
No. And you know that that is the wrong answer because:

$$\frac{d}{dx}\left(\frac{\sin^3(x)}{3}\right) = \sin^2(x)\frac{d}{dx}(\sin x) = \sin^2(x)\cos(x) \not= \cos^2(x)$$

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?

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

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

emilgouliev
integrtion by parts:

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

Last edited by a moderator:
Homework Helper
That seems abit longer and harder than msjds method.
His way easily goes to this:
$$\frac{1}{2} \int 1 dx + \int \cos 2x dx = \frac{1}{2} (x + \int cos 2x dx)$$

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

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

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