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

Question

Cameron Alexander · Accepted Answer

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

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#

Aurora Alvarado · Answer

undefined The indefinite integral of $\sec\left(\frac{x}{2}\right)$ with respect to $x$ is $-\log(\sin(\frac{x}{2}) - 1) + \log(\sin(\frac{x}{2}) + 1) + C$, where $C$ is the constant of integration.