Question

Mostre passo a passo, como encontrar a solução desta integral :

$\int\limits \sqrt{tan\:x} \:dx$

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:

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

Agora, vamos calcular I₂:

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

Podemos voltar à expressão de I:

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

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

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

Answer 2

Calcular a integral indefinida:

$\displaystyle\int\!\sqrt{\tan x}\,dx\qquad\quad\mathbf{(i)}$


Substituição:

$\sqrt{\tan x}=t\quad\Rightarrow\quad\tan x=t^2\quad\Rightarrow\quad\left\{\!\begin{array}{l} x=\arctan(t^2)\\\\ dx=\dfrac{1}{(t^2)^2+1}\cdot 2t\,dt \end{array}\right.$


Substituindo, a integral fica

$\displaystyle\int\!t\cdot \frac{2t}{(t^2)^2+1}\,dt\\\\\\ =\int\!\frac{2t^2}{(t^2)^2+1+2t^2-2t^2}\,dt\quad\longleftarrow\quad\textsf{(completamento de quadrados)}\\\\\\ =\int\!\frac{2t^2}{\big[(t^2)^2+2t^2+1\big]-2t^2}\,dt\\\\\\ =\int\!\frac{2t^2}{(t^2+1)^2-2t^2}\,dt\\\\\\ =\int\!\frac{2t^2}{(t^2+1)^2-\big(\sqrt{2}t\big)^2}\,dt\quad\longleftarrow\quad\textsf{(diferen\c{c}a entre quadrados)}$

$=\displaystyle\int\!\frac{2t^2}{\big[(t^2+1)-\sqrt{2}\,t\big]\cdot \big[(t^2+1)+\sqrt{2}\,t\big]}\,dt\\\\\\ =\int\!\frac{2t^2}{\big(t^2-\sqrt{2}\,t+1\big)\cdot \big(t^2+\sqrt{2}\,t+1\big)}\,dt\qquad\quad\mathbf{(ii)}$


O integrando é uma função racional, com o denominador já fatorado em fatores quadráticos irredutíveis. O próximo passo é decompor o integrando em frações parciais:

$\dfrac{2t^2}{\big(t^2-\sqrt{2}\,t+1\big)\cdot \big(t^2+\sqrt{2}\,t+1\big)}\\\\\\=\dfrac{At+B}{t^2-\sqrt{2}\,t+1}+\dfrac{Ct+D}{t^2+\sqrt{2}\,t+1}\\\\\\=\dfrac{(At+B)\cdot \big(t^2+\sqrt{2}\,t+1\big)+(Ct+D)\cdot \big(t^2-\sqrt{2}\,t+1\big)}{\big(t^2-\sqrt{2}\,t+1\big)\cdot \big(t^2+\sqrt{2}\,t+1\big)}\\\\\\ \footnotesize\begin{array}{l} =\dfrac{(A+C)t^3+\big(\sqrt{2}\,A+B-\sqrt{2}\,C+D\big)t^2+\big(A+\sqrt{2}\,B+C-\sqrt{2}\,D\big)t+(B+D)}{\big(t^2-\sqrt{2}\,t+1\big)\cdot \big(t^2+\sqrt{2}\,t+1\big)} \end{array}$


Por identidade polinomial, devemos ter

$\left\{\! \begin{array}{l} A+C=0\\\\ \sqrt{2}\,A+B-\sqrt{2}\,C+D=2\\\\ A+\sqrt{2}\,B+C-\sqrt{2}\,D=0\\\\ B+D=0 \end{array} \right.$


Resolvendo o sistema acima, encontramos:

$A=\dfrac{1}{\sqrt{2}},~~B=0,~~C=-\,\dfrac{1}{\sqrt{2}},~~D=0.$


Então, a integral $\mathbf{(ii)}$ fica

$=\displaystyle\int\!\bigg(\frac{\frac{1}{\sqrt{2}}\,t}{t^2-\sqrt{2}\,t+1}-\frac{\frac{1}{\sqrt{2}}\,t}{t^2+\sqrt{2}\,t+1}\bigg)dt\\\\\\ =\frac{1}{2\sqrt{2}}\int\!\bigg(\frac{2t}{t^2-\sqrt{2}\,t+1}-\frac{2t}{t^2+\sqrt{2}\,t+1}\bigg)dt\\\\\\ =\frac{1}{2\sqrt{2}}\int\!\bigg(\frac{2t-\sqrt{2}+\sqrt{2}}{t^2-\sqrt{2}\,t+1}-\frac{2t+\sqrt{2}-\sqrt{2}}{t^2+\sqrt{2}\,t+1}\bigg)dt\\\\\\ \footnotesize\begin{array}{l} =\displaystyle\frac{1}{2\sqrt{2}}\int\!\bigg(\frac{2t-\sqrt{2}}{t^2-\sqrt{2}\,t+1}+\frac{\sqrt{2}}{t^2-\sqrt{2}\,t+1}-\frac{2t+\sqrt{2}}{t^2+\sqrt{2}\,t+1}+\frac{\sqrt{2}}{t^2+\sqrt{2}\,t+1}\bigg)dt \end{array}$

$\footnotesize\begin{array}{l} =\displaystyle\frac{1}{2\sqrt{2}}\int\!\frac{2t-\sqrt{2}}{t^2-\sqrt{2}\,t+1}\,dt+\frac{1}{2}\int\!\frac{dt}{t^2-\sqrt{2}\,t+1}-\frac{1}{2\sqrt{2}}\int\!\frac{2t+\sqrt{2}}{t^2+\sqrt{2}\,t+1}\,dt+\frac{1}{2}\int\!\frac{dt}{t^2+\sqrt{2}\,t+1}\\\\ =\displaystyle\frac{1}{2\sqrt{2}}\ln\!\big(t^2-\sqrt{2}\,t+1\big)+\frac{1}{2}\int\!\frac{dt}{\left(t^2-\sqrt{2}\,t+\frac{1}{2} \right )+\frac{1}{2}}-\frac{1}{2\sqrt{2}}\ln\!\big(t^2+\sqrt{2}\,t+1\big)+\frac{1}{2}\int\!\frac{dt}{\left(t^2+\sqrt{2}\,t+\frac{1}{2} \right )+\frac{1}{2}}\\\\ =\displaystyle\frac{1}{2\sqrt{2}}\ln\!\left(\frac{t^2-\sqrt{2}\,t+1}{t^2+\sqrt{2}\,t+1} \right )+\frac{1}{2}\int\!\frac{dt}{\Big(t-\frac{1}{\sqrt{2}}\Big)^{\!2}+\Big(\frac{1}{\sqrt{2}}\Big)^{\!2}}+\frac{1}{2}\int\!\frac{dt}{\Big(t+\frac{1}{\sqrt{2}}\Big)^{\!2}+\Big(\frac{1}{\sqrt{2}}\Big)^{\!2}} \end{array}$

$\footnotesize\begin{array}{l}=\displaystyle\frac{1}{2\sqrt{2}}\ln\!\left(\frac{t^2-\sqrt{2}\,t+1}{t^2+\sqrt{2}\,t+1} \right )+\frac{1}{2}\cdot \frac{~~~1~~~}{\frac{1}{\sqrt{2}}}\arctan\left(\frac{t-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} \right )+\frac{1}{2}\cdot \frac{~~~1~~~}{\frac{1}{\sqrt{2}}}\arctan\left(\frac{t+\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} \right )+C\\\\ =\displaystyle\frac{1}{2\sqrt{2}}\ln\!\left(\frac{t^2-\sqrt{2}\,t+1}{t^2+\sqrt{2}\,t+1} \right )+\frac{1}{\sqrt{2}}\arctan\big(\sqrt{2}\,t-1\big)+\frac{1}{\sqrt{2}}\arctan\big(\sqrt{2}\,t+1\big)+C \end{array}$


Voltando à variável $x,$ finalmente obtemos a resposta final:

$\footnotesize\begin{array}{l} =\displaystyle\frac{1}{2\sqrt{2}}\ln\!\left(\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1} \right )+\frac{1}{\sqrt{2}}\arctan\big(\sqrt{2\tan x}-1\big)+\frac{1}{\sqrt{2}}\arctan\big(\sqrt{2\tan x}+1\big)+C \end{array}$


Bons estudos! :-)


Observação:  Alguns passos foram omitidos na resolução do sistema linear e das integrais que envolvem logaritmos e arco-tangente somente para poupar caracteres (há um limite de 5000 por resposta). Caso tenha alguma dúvida em algum desses passos, comente.