- #1

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## Homework Statement

$1/cosx dx

## Homework Equations

See the scanned paper

## The Attempt at a Solution

See my scanned paper

[PLAIN]http://img812.imageshack.us/img812/2892/img0003001.jpg [Broken]

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- Thread starter Riazy
- Start date

- #1

- 30

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$1/cosx dx

See the scanned paper

See my scanned paper

[PLAIN]http://img812.imageshack.us/img812/2892/img0003001.jpg [Broken]

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- #2

Mark44

Mentor

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1/cos(x) = sec(x). Do you know how to integrate that function?

- #3

LeonhardEuler

Gold Member

- 859

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[tex]\frac{1}{\cos{x}}[/tex]

or

[tex]\frac{1}{\cos^2{x}}[/tex]

because in your first attempt you tried to integrate the first one, while in your second attempt you seem to be integrating the second.

- #4

- 830

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cos^2(x) = 1 -sin^2(x)

- #5

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- #6

- 830

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:rofl::rofl::rofl::rofl::rofl::rofl: LOLMark44:We don't use sec x in sweden, so no i don't

╔(σ_σ)╝: Hmm i could need some more elaboration on that method, I really need to get to solve this, but i won't be able to get in touch with my professor at the moment, and I live 60 kilometers from my uni :/

Well,...

[tex]\int \frac{1}{cosx} \frac{sinx}{sinx} dx[/tex]

Technically speaking, this is not a good idea since sinx or cosx could be zero and you have problems with division by zero. By I assume that the region in question doesn't cause this type of problems.

Now use the substitution u =sinx and show me what you get.

- #7

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[tex] \int \frac{dx}{\cos x} = \int \frac{\cos x}{\cos^2 x} {}dx = \int \frac{d(\sin x)}{1-\sin^2 x } = ... [/tex]

- #8

- 830

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The above suggestions is similar to mine but is actually a better suggestion. ^_^

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