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Asking cos*sec^2, not cos/sec^2

Oh, sorry:
Cosx/Cos^2x = 1/Cosx = Secx

Well actually you'd keep it as 1/Cox, and use integration by parts

wait a second
\[\int cosx\frac 1{cos^2x}dx?\]

Yes.

Sorry for the confusion.. Let's start over:
Cosx*Sec^2x = Cosx/Cos^2x = 1/Cosx, since Secx = 1/Cosx

\[\int\frac 1{cosx}dx\]

\[\int{\cos{x}*\frac{1}{1-\sin^2{x}}}\ dx\]

why did you change it? wouldn't it make more sense to cancel one cosine out?

No

Let u = sin(x)
du = cos(x)dx
\[\int{\frac{du}{1-u^2}}\]

Actually this would be the easiest option, imo:
Cos^2(u) = (1/2)+(Cos(2u))/2

ok let Valpey, finish this, otherwise you'll get confused.

sure

do I multiply the integral by \[\ frac {1+u^2}{1+u^2}\]

\[\frac {1+u^2}{1+u^2}\]

makes sense, thank you!

*\[\frac{1}{2}\Big(\log(1+\sin x)-\log(1-\sin x)\Big)\]

@MathSofiya still want help with\[\int\cos x\sec^2xdx\]?

it seems that everyone here has forgotten the simple trick to this integral...

see there had to be a simpler way

that is the standard approach for this well-known integral

but yes it makes sense now I see it.

I shall memorize this integral =P (honestly I will)
And as always, Thanks Turing!