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Question Number 65200 by rajesh4661kumar@gmail.com last updated on 26/Jul/19

Answered by ajfour last updated on 26/Jul/19

$$I=\int \sqrt{\tan x}dx$$$$let\tan x=t^2\Rightarrow dx=\frac{2tdt}{1+t^2}$$$$\Rightarrow I=\int \frac{2t^{2}dt}{1+t^4}$$$$=\int \frac{(\frac{1}{t^2}+1)dt}{{(-\frac{1}{t}+t)}^2+{\left(\sqrt{2}\right)}^2}+\int \frac{(-\frac{1}{t^2}+1)dt}{{(\frac{1}{t}+t)}^2-{\left(\sqrt{2}\right)}^2}$$$$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln \mid \frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\mid +c$$$$wheret=\sqrt{\tan x}.$$

Answered by zakaria elghaouti last updated on 27/Jul/19

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