Thread: \$1/cosx dx help needed View Single Post

 Quote by Riazy 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.