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]

Last edited by a moderator:

Mentor
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?

Gold Member
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
: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,...
$$\int \frac{1}{cosx} \frac{sinx}{sinx} dx$$
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.

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