# IntegrationWatch

#1
I've been trying to integrate 1/cosx by parts using dv/dx = 1 and u = 1/cos x

I keep getting stuck and was wondering if it is even possible? I know there are other ways to integrate 1/cosx but is it possible by the aforementioned method.

Thanks.
0
3 months ago
#2
so by your method you end up with an even more complicated integral... This is not why we do integration by parts and so should not be used to integrate the function. I know that you probably were told to use it blindly but this is not how you do integrals outside of exam questions. Integration by parts is meant to be used when the new integral becomes simpler!

However this can be useful to find the integral of the more complicated expression ONCE we find the integral in question, which is a nice way to use integration by parts.

really what you should be doing is to "create" terms tat are more familiar to integrate. This is not easy but one can get the hang of it over time.

Here is a hint: multiply both sides by (secx+tanx)/(secx+tanx) where secx is 1/cosx.

Alternatively you can use trig t substitutions but that is maybe not in your syllabus (definitely not if you don't take FM)

Edit: The user below has given a third method
Last edited by TheTroll73; 3 months ago
1
3 months ago
#3
IBP will get you nowhere.
You can multiply top and bottom by cosx giving cosx/((1-sin^2x) then use a u substitution and the difference of two squares to evaluate the integral.
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