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

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In summary, using the substitution u =sinx, the integration can be done to get \int \frac{dx}{\cos x} = \int \frac{\cos x}{\cos^2 x} {}dx = \int \frac{d(\sin x)}{1-\sin^2 x } = \int \frac{\sin x}{\sin^2 x} d\cos x.
  • #1
Riazy
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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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  • #2
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?
 
  • #3
Are you integrating
[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
Try the second method using sinx/sinx then use the substitution u =sinx and the trig identity
cos^2(x) = 1 -sin^2(x)
 
  • #5
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 :/
 
  • #6
Riazy said:
Mark44: We don't use sec x in sweden, so no i don't
:rofl::rofl::rofl::rofl::rofl::rofl: 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,...
[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
Here's a hint

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

1. What does the expression $1/cosx dx mean?

The expression $1/cosx dx represents the indefinite integral of the function 1/cosx, which is also known as the secant function. This means finding the antiderivative of 1/cosx with respect to the variable x.

2. How can I solve the integral $1/cosx dx?

To solve the integral $1/cosx dx, you can use the substitution method by letting u = cosx. This will transform the integral into ∫du/u, which can be easily evaluated using the logarithmic rule.

3. Why is the solution for $1/cosx dx written on a scanned paper?

The solution for $1/cosx dx may be written on a scanned paper to provide a visual aid for understanding the steps and methods used to solve the integral. This can also serve as a reference for those who prefer to see the handwritten solution rather than a typed one.

4. Are there any other ways to solve the integral $1/cosx dx?

Yes, there are several methods to solve the integral $1/cosx dx. Aside from substitution, you can also use the trigonometric identity for secant and tangent, or use integration by parts. It is important to choose the most appropriate method based on the given function and your own understanding.

5. Can I use this solution for $1/cosx dx for other similar integrals?

Yes, the solution for $1/cosx dx can be used as a guide for solving other integrals with a similar form, such as $1/sinx dx or $1/tanx dx. However, it is still important to take note of any variations or specific techniques that may be needed for different functions.

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