Integral of root of $\sec x$

Question

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 .

Answer 1

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.

Answer 2

$$\int_{a}^b \frac{1}{\sqrt{\cos(x)}}dx$$

is defined if $]a,b[\subset ]-\frac{\pi}{2}+2k\pi, \frac{\pi}{2}+2k\pi[$. But you can't calculate it with the usual functions, you'll need "special" functions :

http://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_first_kind

Answer 3

$$\int \sqrt{\sec x} \,dx = \int \frac{1}{\sqrt{\cos x}}\, dx$$; 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.