Need Help with $1/cosx dx? Check Out My Solution | Scanned Paper Inside

[Riazy]

Homework Statement

$1/cosx dx

Homework Equations

See the scanned paper

The Attempt at a Solution

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

 

[Mark44]

An ordinary substitution won't do you any good in this problem.

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

 

[LeonhardEuler]

Are you integrating
\frac{1}{\cos{x}}
or
\frac{1}{\cos^2{x}}
because in your first attempt you tried to integrate the first one, while in your second attempt you seem to be integrating the second.

 

[╔(σ_σ)╝]

Try the second method using sinx/sinx then use the substitution u =sinx and the trig identity
cos^2(x) = 1 -sin^2(x)

 

[Riazy]

Mark44: We don't use sec x in sweden, so no i don't
LeonhardEuler: it's the first one
╔(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 :/

 

[╔(σ_σ)╝]

Mark44: We don't use sec x in sweden, so no i don't

LOL

╔(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,...
\int \frac{1}{cosx} \frac{sinx}{sinx} dx
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.

 

[dextercioby]

Here's a hint

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

 

[╔(σ_σ)╝]

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