∫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 cant i use the chain rule? Can i?

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HallsofIvy
Homework Helper
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.

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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
Homework Helper
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.

1 person
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

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

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