Question

\int\limits {\sqrt{tanx}} \, dx

Answer 1

$\int \sqrt{tgx} dx$

Integral indefinida:

$\int \: \sqrt{ \frac{sen \: x}{cos \: x} } dx \\ \\ \int \: \frac{ \sqrt{sen \: x \: cos \: x} }{cos \: x} dx$

Answer 2

$u^2=tanx\\2udu=sec^2xdx\\ dx=\frac{2udu}{sec^2x} \\ sec^2x=tan^2x+1\\ u^4=tan^2x\\ sec^2x=u^4+1\\ \int\limits {\sqrt{tanx} } \, dx =\int\limits u.\frac{2u}{u^4+1} \, du \\ =\int\limits \frac{\frac{2u^2}{u2} }{\frac{u^4+1}{u^2} } \, du\\ \\ =\int\limits \frac{2}{u^2+\frac{1}{u^2} } \, du\\ \\ u^2+\frac{1}{u^2} = (u+\frac{1}{u} )^2 -2=(u-\frac{1}{u})^2+2\\ \\ \int\limits \frac{1-\frac{1}{u^2} }{(u+\frac{1}{u})^2-2 } \, du + \int\limits \frac{1+\frac{1}{u^2} }{(u-\frac{1}{u})^2+2 } \, du$

$t=u+\frac{1}{u}\\ dt=1-\frac{1}{u^2} du\\ \\ v=u-\frac{1}{u} \\ \\ dv= 1+\frac{1}{u^2} du\\ \\ \int\limits \frac{dt}{t^2-2} \, +\int\limits \frac{dv}{v^2+2} \,$

Sabendo que:

$\int\limits \frac{1}{x^2-a^2} \, dx =\frac{-1}{a} tanh^-1(\frac{x}{a})\\ \\ \int\limits \frac{1}{x^2+a^2} \, dx = \frac{1}{a}tan^-1(\frac{x}{a})$

Temos:

$-\frac{1}{\sqrt{2} } tanh^-1(\frac{\sqrt{tanx} +\frac{1}{\sqrt{tanx} }}{\sqrt{2}} )+\frac{1}{\sqrt{2} } tan^-1(\frac{\sqrt{tanx} -\frac{1}{\sqrt{tanx} }}{\sqrt{2} } )\\ \\ =-\frac{1}{\sqrt{2} } tanh^-1(\frac{\sqrt{tanx}+\sqrt{cotx} }{\sqrt{2} } )+\frac{1}{\sqrt{2} } tan^-1(\frac{\sqrt{tanx}-\sqrt{cotx} }{\sqrt{2} } )+C$