How to integrate "sec(x) dx"?

Question

give answer with explaination reasons.

Answer 1

Someone did it already, so I'll just paste the link here:
http://math2.org/math/integrals/more/sec.htm

Answer 2

This is a classic integration "trick" where we multiply by a very interesting version of 1:

(sec(x)+tan(x))/(sec(x)+tan(x))=1.

Then,
\int sec(x)dx=
\int sec(x) * (sec(x)+tan(x))/
(sec(x)+tan(x))dx=
\int (sec(x)tan(x)+sec^2(x))/
(sec(x)+tan(x))dx

Now, notice that the derivative of the denominator is the numerator because d/dx (sec(x))=sec(x)tan(x) and d/d (tan(x))=sec^2(x). So, we let

u=sec(x)+tan(x)

Then, du=(sec(x)tan(x)+sec^2(x))dx.

Subsitution into the original integral gives us

\int (sec(x)tan(x)+sec^2(x))/
(sec(x)+tan(x))dx=
\int 1/u du=
ln |u|+C=
=ln|sec(x)+tan(x)|+C

You can use the same idea to integrate csc(x), too.