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Discussion Response Post to: find this integral

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4 Jun 2008 21:13:57 IST

Subject: Re:find this integral

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Greatdreams (3282)

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$\sqrt{tan x} , take\;tan \;x = t^2 \\d(tan x ) = d (t^2) \\\Rightarrow sec^2x \; dx = 2t\; dt\\ \int\sqrt{tan x} \;dx = \int \sqrt{t^2} \;2t/(1+t^4) dt$

$= \int 2t^2/1+t^4 dt \\ = \int 2 / t^2+1/t^2 dt\\=\int (1+1/t^2)+(1-1/t^2)/(t^2+1/t^2 \; dt$

$= \int 1+1/t^2 /t^2+1/t^2 dt + \int 1-1/t^2/(t^2+1/t^2)dt$

$=\int 1/(t-1/t)^2+2 \; \;d(t-1/t) + \int 1/(t+1/t)^2 d(t+1/t)$

$= \int du/(u^2 + (\sqrt{2})^2 + \int dv/v^2-(\sqrt{2})^2 dv$

$where \; u = t-1/t \;and\; v=t+1/t$

$=1/\sqrt{2}tan^{-1}(u/\sqrt{2}) + 1/2\sqrt{2}\;log [ v-\sqrt{2}/v+\sqrt{2}] + c$

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