Question

How do you find the indefinite integral of `int sec(x/2)`?

Answer 1

`2ln|sec(x/2)+tan(x/2)|+C`

Explanation:

With regards to integrating secants, there's really no way to do it without manipulating the integrand. So, we'll do the following

`sec(x/2)=sec(x/2)*(sec(x/2)+tan(x/2))/(sec(x/2)+tan(x/2))` (Multiplying by one)

Thus,

`intsec(x/2)dx=intsec(x/2)*(sec(x/2)+tan(x/2))/(sec(x/2)+tan(x/2))dx`

Now, we can say `u=sec(x/2)+tan(x/2)`, referring to the denominator.

Thus,

`du=1/2sec(x/2)tan(x/2)+1/2sec^2(x/2)dx`

Let's get rid of the fractions on the right since our integrand has no fractions outside of the arguments of the trigonometric functions:

`2du=sec(x/2)tan(x/2)+sec^2(x/2)dx`

At first glance, it doesn't look like this shows up in our integral, but multiplying out the numerator yields:

`int(sec^2(x/2)+sec(x/2)tan(x/2))/(sec(x/2)+tan(x/2))dx`

So, this substitution is indeed valid, and it gives us

`2int(du)/u=2ln|u|+C=2ln|sec(x/2)+tan(x/2)|+C`