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Too Easy To Calculate

sos440 2012. 4. 23. 21:20

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!



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