I really like this, because it doesn't rely on just "knowing" the trick. Doing these steps is something that, conceivably, I might think to do & it's great to see the transformation at the end. Brilliant presentation, thank you!

you're a genius man i definitely rate you five stars...

Good job buddy!

Nice! 😃

Thank you. That's the best explanation I have seen

Tks, professor! You have a new sobscriber 🤗

"REALLY GOOD!!!!!"

Yes I learned a lot

...Newton, Because of your boundless enthusiasm during the presentation, sparks fly off the top right corner of the blackboard at about time 7:43, taking a (du) of the indefinite integral ( (1/2)INT(1/(1-u)) )with them! (lol). Great presentation with a lot of entertaining mathematical manipulations...  Thank you,    Jan-W

This could also be written as
arctanh sinx + c
Because
Integral[ du / (1 - u^2 )] = arctanh u

Nice, sir.  S Chitrai Kani

Okdok you bet

...Good day to you Newton, I hope you feel fine. I just wanted to share with you the following from my humble mathematical past, given that you indicated that this indefinite integral (sec(x)) can be solved in several ways. I have kept all my math notebooks from the past, and so you see that they can still be useful in the present, also for other interested people. I believe I already told you that I had a lazy (read "efficient") yet strict and fair math teacher. I therefore found the following solution method in my writing: Instead of multiplying the integrand sec(x) by cos(x)/cos(x), multiply it by (sec(x) + tan(x))/(sec(x) + tan(x)), then distribute the expression of the numerator, and for the u-substitution take u = sec(x) + tan(x) --> du = (numerator expression) times dx. So the new integrand becomes 1/u. The rest is self - explanatory, namely that after integrating 1/u, directly after back substitution from the u-world to the x-world you get exactly the same solution as ln I sec(x) + tan(x) I + C. A lazy (again read efficient) math teacher can have advantages (lol). However I must admit that this method comes across as more of a trick (bad for MY own understanding and memory), while I think your method presented seems more logical, and will therefore remain in my memory longer! Newton, if you have read my story, I thank you for this...   Take good care,     Jan-W

you have used the partial fraction decomposition, before that step, why don't you manipulate the function by multiplying by (-2/-2) so that the numerator is the derivative of the denominator and factor out the -ve half out?

We could just multiply and divide with secx+tanx. THe numerator would be sec^2 x + secxtanx, while the denominator would be  secx+tan x. We have the differentiation of the denominator on the numerator, and hence, using u substitution, we'll finally reach ln|secx+tanx| + C as the answer.

We have  Secx =(secx tanx +(secx)2)/(tanx +secx )  Then the integration of Secx is ln l secx +tanx l +c