Question

Calcule $\int\limits { \sqrt{tan(x)} \, dx$

Answer 1

Vamos fazer uma substituição:

$y=\sqrt{\tan x}=(\tan x)^{\frac{1}{2}}\\\\ dy=\dfrac{1}{2}\cdot(\tan x)^{-\frac{1}{2}}\cdot\sec^2x\,dx\\\\ 2dy=\dfrac{1}{y}\cdot(\tan^2x+1)dx\\\\ 2ydy=(y^4+1)dx\\\\ dx=\dfrac{2y}{y^4+1}dy$

Usando a substituição na integral dada:

$I=\displaystyle\int\sqrt{\tan x}\,dx\\\\ I=\displaystyle\int y\cdot\dfrac{2y}{y^4+1}dy\\\\ I=\displaystyle\int\dfrac{2y^2}{y^4+1}dy$

Vamos fatorar o polinômio (y⁴+1) usando o Lema de Gauss. Podemos ver que não ha raízes reais, logo vamos diretamente tentar fatorar em dois polinômios de grau 2:

$y^4+1=(y^2+ay+1)(y^2+cy+1)\\\\ y^4+1=y^4+(a+c)y^3+(ac+2)y^2+(a+c)y+1\\\\ \begin{cases}a+c=0\to c=-a\\ac+2=0\to ac=-2\to -a^2=-2\to a=\sqrt2,\,c=-\sqrt2\end{cases}\\\\ \Longrightarrow (y^4+1)=(y^2+\sqrt2y+1)(y^2-\sqrt2y+1)$

Agora podemos usar o método de frações parciais. Olhando agora para a fração de dentro da integral:

$\dfrac{2y^2}{y^4+1}=\dfrac{Ay+B}{y^2+\sqrt2y+1}+\dfrac{Cy+D}{y^2-\sqrt2y+1}\\\\=\dfrac{(A+C)y^3+(-\sqrt2A+B+\sqrt2C+D)y^2+(A-\sqrt2B+C+\sqrt2D)y+(B+D)}{y^4+1}\\\\ \Longrightarrow\begin{cases}A+C=0\to A=-C\\-\sqrt2(A-C)+(B+D)=2\to-\sqrt2\cdot2A+0=2\to\boxed{A=-\frac{1}{\sqrt2}},\,\boxed{C=\frac{1}{\sqrt2}}\\(A+C)+\sqrt2(D-B)=0\to 0+\sqrt2(D-B)=0\to\boxed{B=D=0}\\B+D=0\end{cases}$

Usando o que foi feito acima:

$I=\displaystyle\int\dfrac{-\frac{1}{\sqrt2}y}{y^2+\sqrt2y+1}dy+\displaystyle\int\dfrac{\frac{1}{\sqrt2}y}{y^2-\sqrt2y+1}dy\\\\\\I=-\dfrac{1}{\sqrt2} \underbrace{\displaystyle\int\dfrac{y}{y^2+\sqrt2y+1}dy}_{I_1}+\dfrac{1}{\sqrt2} \underbrace{\displaystyle\int\dfrac{y}{y^2-\sqrt2y+1}dy}_{I_2}$

Resolveremos cada uma das integrais destacadas separadamente. Começando por I₁:

$I_1=\displaystyle\int\dfrac{y}{y^2+\sqrt2+1}dy\\\\ 2I_1=\displaystyle\int\dfrac{2y}{y^2+\sqrt2y+1}dy=\displaystyle\int\dfrac{2y-\sqrt2+\sqrt2}{y^2+\sqrt2y+1}dy\\\\ 2I_1=\displaystyle\int\dfrac{2y+\sqrt2}{y^2+\sqrt2y+1}dy-\displaystyle\int\dfrac{\sqrt2}{y^2+\sqrt2y+1}dy\\\\ 2I_1=\ln|y^2+\sqrt2y+1|-2\sqrt2\displaystyle\int\dfrac{1}{(y\sqrt2+1)^2+1}dy\\\\ 2I_1=\ln|y^2+\sqrt2y+1|-2\sqrt2\cdot\dfrac{1}{\sqrt2}\arctan(y\sqrt2+1)\\\\\boxed{I_1=\frac{1}{2}\ln|y^2+\sqrt2y+1|-\arctan(y\sqrt2+1)}$

Agora, vamos calcular I₂:

$I_2=\displaystyle\int\dfrac{y}{y^2-\sqrt2+1}dy\\\\ 2I_2=\displaystyle\int\dfrac{2y}{y^2-\sqrt2y+1}dy=\displaystyle\int\dfrac{2y-\sqrt2+\sqrt2}{y^2+\sqrt2y+1}dy\\\\ 2I_2=\displaystyle\int\dfrac{2y-\sqrt2}{y^2-\sqrt2y+1}dy+\displaystyle\int\dfrac{\sqrt2}{y^2-\sqrt2y+1}dy\\\\ 2I_2=\ln|y^2-\sqrt2y+1|+2\sqrt2\displaystyle\int\dfrac{1}{(y\sqrt2-1)^2+1}dy\\\\ 2I_2=\ln|y^2-\sqrt2y+1|+2\sqrt2\cdot\dfrac{1}{\sqrt2}\arctan(y\sqrt2-1)\\\\\boxed{I_2=\frac{1}{2}\ln|y^2-\sqrt2y+1|+\arctan(y\sqrt2-1)}$

Podemos voltar à expressão de I:

$I=-\dfrac{I_1}{\sqrt2}+\dfrac{I_2}{\sqrt2}\\\\ I=-\dfrac{\frac{1}{2}\ln|y^2+\sqrt2y+1|-\arctan(y\sqrt2+1)}{\sqrt2}+\\+\dfrac{\frac{1}{2}\ln|y^2-\sqrt2y+1|+\arctan(y\sqrt2-1)}{\sqrt2}\\\\ I=-\frac{1}{2\sqrt2}\ln|y^2+\sqrt2y+1|+\frac{1}{\sqrt2}\arctan(y\sqrt2+1)+\\+\frac{1}{2\sqrt2}\ln|y^2-\sqrt2y+1|+\frac{1}{\sqrt2}\arctan(y\sqrt2-1)$

Agora, voltando à expressão em função de x, chegamos finalmente à resposta final:

$I=-\frac{1}{2\sqrt2}\ln|y^2+\sqrt2y+1|+\frac{1}{\sqrt2}\arctan(y\sqrt2+1)+\\+\frac{1}{2\sqrt2}\ln|y^2-\sqrt2y+1|+\frac{1}{\sqrt2}\arctan(y\sqrt2-1)\\\\\\I=-\frac{1}{2\sqrt2}\ln|(\sqrt{\tan x})^2+\sqrt2(\sqrt{\tan x})+1|+\frac{1}{\sqrt2}\arctan((\sqrt{\tan x})\sqrt2+1)\\+\frac{1}{2\sqrt2}\ln|(\sqrt{\tan x})^2-\sqrt2(\sqrt{\tan x})+1|+\frac{1}{\sqrt2}\arctan((\sqrt{\tan x})\sqrt2-1)\\\\\\\boxed{\boxed{I=-\frac{1}{2\sqrt2}\ln|\tan x+\sqrt{2\tan x}+1|+\frac{1}{\sqrt2}\arctan(\sqrt{2\tan x}+1)+}}\\\boxed{\boxed{+\frac{1}{2\sqrt2}\ln|\tan x-\sqrt{2\tan x}+1|+\frac{1}{\sqrt2}\arctan(\sqrt{2\tan x}-1)+C}}$