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 [tex]dy/dt= (dy/dx)(dx/dt)[/tex]. 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, [itex]\int u^2 dx= (1/3)u^3/u'+ C[/itex] 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 [itex] u [/itex], you want to integrate with respect to [itex] u [/itex], 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, [itex]\int\cos{(x)}\cos{(x)}dx[/itex]. Try integrating this

[itex] (fg)' = f'g+ g'f [/itex]