# $$\int \sqrt{\tan x}\ dx$$
$使用代換積分法(u-substitution)$

$\begin{split}
\\令\ u &= \sqrt{\tan x} \\\\
u^{2} &= \tan x \\\\
2u \ du &= \sec^{2}x\ dx \\\\
dx &= \dfrac{2u\ du}{\sec^{2}x} \\\\
\end{split}$

然而$\sec^{2}x$並不是$u$的函數
因此要將$\sec^{2}x$轉為$u$的函數
所以利用
$\tan x = u^{2}$
$\tan^{2}x = u^{4}$
又
$\tan^{2}x = \sec^{2}-1$
所以
$\sec^{2}-1 = u^{4}$
$\sec^{2} = u^{4}+1$

$dx = \dfrac{2u\ du}{u^{4}+1}$

>代入
$\sqrt{\tan x}=u$<br />
$dx = \dfrac{2u\ du}{u^{4}+1}$

$\quad \displaystyle \int {\sqrt{\tan x}\ dx}$

$\begin{align}
&=\displaystyle \int {u \cdot \dfrac{2u}{u^{4}+1}\ du}\\
\\ &=\displaystyle \int {\dfrac{2u^{2}}{u^{4}+1}\ du}\\
\\ &=\displaystyle \int {\dfrac{\dfrac{2u^{2}}{u^{2}}}{\dfrac{u^{4}+1}{u^{2}}}\ du}\\
\\&=\displaystyle \int {\dfrac{2}{u^{2}+\dfrac{1}{u^{2}}}\ du}
\end{align}$

>$\begin{align}
u^{2}+\dfrac{1}{u^{2}}
&= (u)^{2}+(\dfrac{1}{u})^{2}\\
\\ &= (u)^{2}+2(u)(\dfrac{1}{u})+(\dfrac{1}{u})^{2}-2(u)(\dfrac{1}{u})\\
\\ &=(u+\dfrac{1}{u})^{2}-2
\end{align}$
$\begin{align}
\\\dfrac{d}{du}(u+\dfrac{1}{u})
&=1-\dfrac{1}{u^{2}}
\end{align}$

>$\begin{align}
u^{2}+\dfrac{1}{u^{2}}
&= (u)^{2}+(\dfrac{1}{u})^{2}\\
\\ &= (u)^{2}-2(u)(\dfrac{1}{u})+(\dfrac{1}{u})^{2}+2(u)(\dfrac{1}{u})\\
\\ &=(u-\dfrac{1}{u})^{2}+2
\end{align}$
$\begin{align}
\\\dfrac{d}{du}(u-\dfrac{1}{u})
=1+\dfrac{1}{u^{2}}
\end{align}$

拆成兩部分
$\begin{align}
\\ &=\displaystyle \int {\dfrac{1-\dfrac{1}{u^{2}}}{(u+\dfrac{1}{u})^{2}-2}+\dfrac{1+\dfrac{1}{u^{2}}}{(u-\dfrac{1}{u})^{2}+2}\ du}
\end{align}$
>令
>$t=u+\dfrac{1}{u}$
>$w=u-\dfrac{1}{u}$
>$dt=(1-\dfrac{1}{u^{2}})\ du$
>$dw=(1+\dfrac{1}{u^{2}})\ du$

代換
$\begin{align}
\\ &=\displaystyle \int \dfrac{1}{t^{2}-2}\ dt\ +\ \displaystyle \int \dfrac{1}{w^{2}+2}\ dw
\end{align}$
>$\displaystyle \int \dfrac{1}{x^{2}-a^{2}}\ dx=\dfrac{-1}{a}\ \tanh^{-1}(\dfrac{x}{a})$
>
>$\displaystyle \int \dfrac{1}{x^{2}+a^{2}}\ dx=\dfrac{1}{a}\ \tan^{-1}(\dfrac{x}{a})$

$\begin{align}
&=\dfrac{-1}{\sqrt{2}}\ \tanh^{-1}(\dfrac{t}{\sqrt{2}})+\dfrac{1}{\sqrt{2}}\ \tan^{-1}(\dfrac{w}{\sqrt{2}})\\
\\ &=\dfrac{-1}{\sqrt{2}}\ \tanh^{-1}(\dfrac{u+\dfrac{1}{u}}{\sqrt{2}})+\dfrac{1}{\sqrt{2}}\ \tan^{-1}(\dfrac{u-\dfrac{1}{u}}{\sqrt{2}})\\
\end{align}$
>代入
>$\begin{align}
>u&=\sqrt{\tan x}
>\\\dfrac{1}{u} &=\sqrt{\cot x}
>\end{align}$

$=\dfrac{-1}{\sqrt{2}}\ \tanh^{-1}(\dfrac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{2}})+\dfrac{1}{\sqrt{2}}\ \tan^{-1}(\dfrac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}})+C$