- #1

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## Homework Statement

∫cos

^{2}x dx

## The Attempt at a Solution

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

cos2x+sin2x=1

cos

^{2}x-sin

^{2}x=cos2x

cos

^{2}x=(1+cos2x)/2

But why cant i use the chain rule? Can i?

- Thread starter johann1301
- Start date

- #1

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∫cos

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

cos2x+sin2x=1

cos

cos

But why cant i use the chain rule? Can i?

- #2

HallsofIvy

Science Advisor

Homework Helper

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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.

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.

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- #3

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u=cosxWhy would you expect to be able to use it to integrate?

therefore...

∫(cos

chain rule reversed?

- #4

HallsofIvy

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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.u=cosx

therefore...

∫(cos^{2}x)dx = ∫(u^{2})dx = (1/3)u^{3}/u' + C = (1/3)cosx^{3}/(cosx)' + C = (1/3)cosx^{3}/(-sinx) + C

chain rule reversed?

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]

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