This problem is a challenge question given to me by my friend. He asked me to find
$$\displaystyle{I=\int\sqrt{\sec x}}\ dx$$
What I tried:
Let $\sec x=u$. So $$du = \sec x\tan x\ dx=\sec x\sqrt{\sec ^2 x-1}\ dx=u\sqrt{u^2-1}\ dx$$$$\Rightarrow\ \ \ \ \ dx = \frac{du}{u\sqrt{u^2-1}}$$Hence,$$I=\int\frac{\sqrt u}{u\sqrt{u^2-1}}du = \int\frac1{\sqrt{u\left(u^2-1\right)}}du$$From this point on I couldn't find any good substitutions. So I tried doing it by parts.$$\begin{align} I&=\int\frac1{\sqrt u}\cdot\frac1{\sqrt{u^2-1}}du\\ &=\frac1{\sqrt{u^2-1}}\int\frac1{\sqrt u}du -\int\left[\frac{-u}{(u^2-1)^{3/2}}\int\frac{1}{\sqrt u}du\right]du\\ &=\frac{2\sqrt u}{\sqrt{u^2-1}}+\int\frac{2u^{3/2}}{(u^2-1)^{3/2}}du\end{align}$$
However this has got even more complicated and I really have no hint on how to do it. Please help me if you know the correct method.
Thanks for the attention.
$\endgroup$ 21 Answer
$\begingroup$From trigonometric half angle formula we have $\cos x=1-2\sin^2\left(\dfrac x2\right)$ and we get
$$\int\sqrt{\sec x}\ dx=\int\dfrac{1}{\sqrt{\cos x}}\ dx=\int\dfrac{1}{\sqrt{1-2\sin^2\left(\dfrac x2\right)}}\ dx = 2F\bigg(\dfrac x2\bigg|2\bigg )+C$$
Where $ F(x|k) $ is an Elliptic Integral of First Type
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