TITLE: Integral of root of $\sec x$
QUESTION [6 upvotes]: How to evaluate the integral 
$$\int \sqrt{\sec x} \, dx$$
I read that its not defined.
But why is it so ? Does it contradict some basic rules ?
Please clarify it .

REPLY [6 votes]: First notice that $\cos x = 1 - 2\sin^2 \Big(\frac{x}{2}\Big)$ then 
$$\int \frac{1}{\sqrt{\cos x}} \, dx  = \int \frac{1}{\sqrt{1 - 2\sin^2 \Big(\frac{x}{2}\Big)}} \, dx = \color{red}{2F\Big(\left.\frac{x}{2}\right\vert 2\Big)} + C$$
where $F(\left.x\right\vert m)$ is an Elliptic Integral of First kind.

REPLY [4 votes]: $$\int \sqrt{\sec x} \,dx = \int \frac{1}{\sqrt{\cos x}}\, dx$$; Now substituting $u = \cos x$ then it follows $$du = - \sin x \, dx = - \sqrt{1-u^2} dx.$$ Hence:
$$\int \sqrt{\sec x} \, dx = - \int \frac{1}{\sqrt{u(1-u^2)}}du$$ This integral is an elliptic integral.