Integrating cos^2x with the Chain Rule: Explanation and Example

[johann1301]

Homework Statement

∫cos²x dx

The Attempt at a Solution

I know the answer, and i know how to get there using:

cos2x+sin2x=1
cos²x-sin²x=cos2x
cos²x=(1+cos2x)/2

But why can't i use the chain rule? Can i?

 

[HallsofIvy]

I am puzzled by your question. The chain rule is a method for differentiation. Why would you expect to be able to use it to integrate?

Integration is the opposite of differentiation. Perhaps you are asking "why can't I reverse the chain rule to integrate this?".

The chain rule says that if x is a function of some variable, t, and y is a function of x, then dy/dt= (dy/dx)(dx/dt). To use it we have to multiply by dx/dt. To do that in reverse, which is "substitution", the "dx/dt" term has to already be in the integral, you can't just insert it.

 

[johann1301]

Why would you expect to be able to use it to integrate?

u=cosx

therefore...

∫(cos²x)dx = ∫(u²)dx = (1/3)u³/u' + C = (1/3)cosx³/(cosx)' + C = (1/3)cosx³/(-sinx) + C

chain rule reversed?

 

[HallsofIvy]

Your second equality, \int u^2 dx= (1/3)u^3/u'+ C is incorrect for precisely the reasons I explained in my first response.

 

[matineesuxxx]

u=cosx

therefore...

∫(cos²x)dx = ∫(u²)dx = (1/3)u³/u' + C = (1/3)cosx³/(cosx)' + C = (1/3)cosx³/(-sinx) + C

chain rule reversed?

You must also consider the variable of integration. Since you've made a substitution u, you want to integrate with respect to u, and as HallsofIvy said, this is moot because of the chain rule.

Ever heard of integration by parts? You should rather think of this as a product, \int\cos{(x)}\cos{(x)}dx. Try integrating this

(fg)' = f'g+ g'f