Too Easy To Calculate

The purpose of this posting is to encourage this blog not to suffer from stagnation, though I'm not sure whether it will work or not.

Here, we introduce a very easy calculation, though in a slightly unfamiliar fashion.

\begin{align*} \int\sqrt{\tan x}\;dx &=\int\frac{2t^2}{t^4+1}\;dt\qquad(t=\sqrt{\tan x})\\ &=\int\frac{2}{t^2+t^{-2}}\;dt\\ &=\int\left(\frac{1-t^{-2}}{\left(t+t^{-1}\right)^2-2}+\frac{1+t^{-2}}{\left(t-t^{-1}\right)^2+2}\right)\;dt\\ &=\int\frac{du}{u^2-2}+\int\frac{dv}{v^2+2}\qquad(u=t+t^{-1}, v=t-t^{-1})\\ &=\frac{1}{\sqrt{2}}\left[\tan^{-1}\left(\frac{v}{\sqrt{2}}\right)-\tanh^{-1}\left(\frac{u}{\sqrt{2}}\right)\right]+C\\ &=\frac{1}{\sqrt{2}}\left[\tan^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)-\tanh^{-1}\left(\frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{2}}\right)\right]+C. \end{align*}

…and what I have to do next is just go to bed and sleep! Goodnight everyone!