Exercise:
Integrate, $\int{\sqrt{\tan
x}\,dx}$
Solution:
Let ${{z}^{2}}=\tan x$$\Rightarrow 2zdz={{\sec }^{2}}x\,dx$
So $\int{\sqrt{\tan
x}\,dx}=\int{z\,\frac{2zdz}{{{\sec }^{2}}x}}=\int{\frac{2{{z}^{2}}}{{{\sec
}^{2}}x}dz}$ but ${{\sec }^{2}}x=1+{{\tan }^{2}}x$
$\Rightarrow
\int{\sqrt{\tan x}dx}=\int{\frac{2{{z}^{2}}dz}{1+{{\tan
}^{2}}x}=\int{\frac{2{{z}^{2}}dz}{1+{{z}^{4}}}}}=2\int{\frac{{{z}^{2}}}{1+{{z}^{4}}}dz}$
$=\int{\frac{2{{z}^{2}}+1-1}{1+{{z}^{4}}}dz}$
$=\int{\frac{{{z}^{2}}+1+{{z}^{2}}-1}{1+{{z}^{4}}}dz}$
$=\int{\frac{{{z}^{2}}+1}{{{z}^{4}}+1}dz+\int{\frac{{{z}^{2}}-1}{{{z}^{4}}+1}dz}}$
$\Rightarrow \int{\frac{{{z}^{2}}+1}{{{z}^{4}}+1}dz}=\int{\frac{{{z}^{2}}\left(
1+\frac{1}{{{z}^{2}}} \right)}{{{z}^{2}}\left( {{z}^{2}}+\frac{1}{{{z}^{2}}}
\right)}dz}$ $=\int{\frac{1+\frac{1}{{{z}^{2}}}}{{{z}^{2}}+\frac{1}{{{z}^{2}}}}dz}$
Let $u=z-\frac{1}{z}\Leftrightarrow
du=1+\frac{1}{{{z}^{2}}}dz$ & ${{u}^{2}}={{z}^{2}}+\frac{1}{{{z}^{2}}}-2\frac{1}{z}z={{z}^{2}}+\frac{1}{{{z}^{2}}}$
$\Rightarrow {{u}^{2}}+2={{z}^{2}}+\frac{1}{{{z}^{2}}}$
So $\int{\frac{1+\frac{1}{{{z}^{2}}}}{{{z}^{2}}+\frac{1}{{{z}^{2}}}}dz}=\int{\frac{du}{{{u}^{2}}+2}}$ \( =\int{\frac{du}{{{u}^{2}}+{{\left( \sqrt{2}
\right)}^{2}}}=\int{\frac{\sqrt{2}du}{{{\left( \sqrt{2} \right)}^{2}}\left[
{{\left( \frac{u}{\sqrt{2}} \right)}^{2}}+1 \right]}=\frac{1}{\sqrt{2}}\int{\frac{du}{{{\left(
\frac{u}{\sqrt{2}} \right)}^{2}}+1}}}}\) \(=\frac{1}{\sqrt{2}}\arctan \left(
\frac{u}{\sqrt{2}} \right)+c\)
thus $\int{\frac{{{z}^{2}}+1}{1+{{z}^{4}}}dz}=\frac{1}{2}\arctan
\left( \frac{{{z}^{2}}-1}{z\sqrt{2}} \right)+c=\frac{1}{2}\arctan \left(
\frac{\tan x-1}{\sqrt{2\tan x}} \right)+c$
$\int{\frac{{{z}^{2}}-1}{{{z}^{4}}+1}dz}=\int{\frac{{{z}^{2}}\left(
1-\frac{1}{{{z}^{2}}} \right)}{{{z}^{2}}\left( {{z}^{2}}+\frac{1}{{{z}^{2}}}
\right)}dz}$ , Let $w=z+\frac{1}{z}\Rightarrow dw=\left( 1-\frac{1}{{{z}^{2}}}
\right)dz$ & ${{w}^{2}}={{z}^{2}}+\frac{1}{{{z}^{2}}}+2$
So $\int{\frac{{{z}^{2}}-1}{{{z}^{4}}+1}dz}=\int{\frac{dw}{{{w}^{2}}-2}}=\int{\frac{dw}{\left(
w-\sqrt{2} \right)\left( w+\sqrt{2} \right)}}$
So $\frac{1}{{{w}^{2}}-2}=\frac{A}{w-\sqrt{2}}+\frac{B}{w+\sqrt{2}}=\frac{A\left(
w+\sqrt{2} \right)+B\left( w-\sqrt{2} \right)}{{{w}^{2}}-2}=\frac{w\left( A+B
\right)+\sqrt{2}\left( A-B \right)}{{{w}^{2}}-2}$
Thus we get $A-B=\frac{1}{\sqrt{2}}\,\,\,\And
\,\,A+B=0$ i.e $A=-B$ and $B=-\frac{1}{2\sqrt{2}}\,\,,\,\,\,A=\frac{1}{2\sqrt{2}}$
So $\int{\frac{dw}{{{w}^{2}}-2}=\frac{1}{2\sqrt{2}}\int{\frac{dw}{w-\sqrt{2}}-\frac{1}{2\sqrt{2}}\int{\frac{dw}{w+\sqrt{2}}=\frac{1}{2\sqrt{2}}\ln
\left| \frac{w-\sqrt{2}}{w+\sqrt{2}} \right|+c}}}$
Thus $\int{\frac{{{z}^{2}}-1}{{{z}^{4}}+1}dz}=\frac{1}{2\sqrt{2}}\ln
\left| \frac{{{z}^{2}}+1-z\sqrt{2}}{{{z}^{2}}+1+z\sqrt{2}} \right|+c$ $=\frac{1}{2\sqrt{2}}\ln
\left| \frac{\tan x+1-\sqrt{2\tan x}}{\tan x+1+\sqrt{2\tan x}} \right|+c$
$\therefore \int{\sqrt{\tan x}\,dx}=\frac{1}{2}\arctan
\left( \frac{\tan x-1}{\sqrt{2\tan x}} \right)+\frac{1}{2\sqrt{2}}\ln \left|
\frac{\tan x+1-\sqrt{2\tan x}}{\tan x+1+\sqrt{2\tan x}} \right|+c$