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\[\begin{align}
& I=\int{\sqrt{\tan x}\,dx} \\
& Solution\,\,I=\int{\frac{\sqrt{\tan x}\cdot {{\sec }^{2}}x}{{{\sec }^{2}}x}\,dx} \\
& Let\,\,\tan x={{u}^{2}}\,\Rightarrow \,{{\sec }^{2}}x\,dx=2u\,du \\
& I=2\int{\frac{{{u}^{2}}}{{{u}^{2}}+1}\,du} \\
& =2\int{\frac{{{u}^{2}}+1-1}{{{u}^{2}}+1}\,du}=2\int{\left( 1-\frac{1}{{{u}^{2}}+1} \right)}\,du \\
& =2\left( u-{{\tan }^{-1}}u \right)+C \\
& =2\left( \sqrt{\tan x}-{{\tan }^{-1}}\left( \sqrt{\tan x} \right) \right)+C \\
\end{align}\]
; 09-22-2016 03:00 PM
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