Question

Pesso ajuda gente. A que é igual a Integral de 1/cos^3(x)

Answer 1

Calcular a integral indefinida

     $\displaystyle\int\dfrac{1}{\cos^3 x}\,dx\\\\\\ =\int\dfrac{1}{\cos^4 x}\cdot \cos x\,dx\\\\\\ =\int\dfrac{1}{(\cos^2 x)^2}\cdot \cos x\,dx\\\\\\ =\int\dfrac{1}{(1-\mathrm{sen^2\,}x)^2}\cdot \cos x\,dx$


Faça a seguinte mudança de variável:

     $\mathrm{sen\,}x=u\quad\Longrigharrow\quad \cos x\,dx=du$


e a integral fica

     $\displaystyle=\int\dfrac{1}{(1-u^2)^2}\,du\\\\\\ =\int\dfrac{1}{\big[(1-u)(1+u)\big]^2}\,du\\\\\\ =\int\dfrac{1}{(1-u)^2(1+u)^2}\,du\qquad\quad\mathsf{(i)}$


Vamos decompor o integrando em frações parciais.

     $\dfrac{1}{(1-u)^2(1+u)^2}=\dfrac{A}{1-u}+\dfrac{B}{(1-u)^2}+\dfrac{C}{1+u}+\dfrac{D}{(1+u)^2}$

     $\dfrac{1}{(1-u)^2(1+u)^2}=\dfrac{A(1-u)(1+u)^2+B(1+u)^2+C(1-u)^2(1+u)+D(1-u)^2}{(1-u)^2(1+u)^2}$


Igualando apenas os numeradores:

     $1=A(1-u)(1+u)^2+B(1+u)^2+C(1-u)^2(1+u)+D(1-u)^2$


Fazendo u = 1, ficamos com

     $1=A(1-1)(1+1)^2+B(1+1)^2+C(1-1)^2(1+1)+D(1-1)^2\\\\ 1=B\cdot 2^2\\\\ 1=A\cdot 0\cdot 2^2+B\cdot 2^2+C\cdot 0^2\cdot 2+D\cdot 0^2\\\\ 1=B\cdot 4\\\\ B=\dfrac{1}{4}$


Fazendo u = -1, ficamos com

     $1=A(1+1)(1-1)^2+B(1-1)^2+C(1+1)^2(1-1)+D(1+1)^2\\\\ 1=B\cdot 2^2\\\\ 1=A\cdot 2\cdot 0^2+B\cdot 0^2+C\cdot 2^2\cdot 0+D\cdot 2^2\\\\ 1=D\cdot 4\\\\ D=\dfrac{1}{4}$


Então, a igualdade fica

     $1=A(1-u)(1+u)^2+\dfrac{1}{4}(1+u)^2+C(1-u)^2(1+u)+\dfrac{1}{4}(1-u)^2$


Para u = 0, obtemos

     $1=A(1-0)(1+0)^2+\dfrac{1}{4}(1+0)^2+C(1-0)^2(1+0)+\dfrac{1}{4}(1-0)^2\\\\\\ 1=A\cdot 1\cdot 1^2+\dfrac{1}{4}\cdot 1^2+C\cdot 1^2\cdot 1+\dfrac{1}{4}\cdot 1^2\\\\\\ 1=A+\dfrac{1}{4}+C+\dfrac{1}{4}\\\\\\ A+C=1-\dfrac{1}{4}-\dfrac{1}{4}\\\\\\ A+C=\dfrac{1}{2}$


Para u = 2, obtemos

     $1=A(1-2)(1+2)^2+\dfrac{1}{4}(1+2)^2+C(1-2)^2(1+2)+\dfrac{1}{4}(1-2)^2\\\\\\ 1=A\cdot (-1)\cdot 3^2+\dfrac{1}{4}\cdot 3^2+C\cdot (-1)^2\cdot 3+\dfrac{1}{4}\cdot (-1)^2\\\\\\ 1=-9A+\dfrac{9}{4}+3C+\dfrac{1}{4}\\\\\\ -9A+3C=1-\dfrac{9}{4}-\dfrac{1}{4}\\\\\\ -9A+3C=-\dfrac{6}{4}\\\\\\ -9A+3C=-\dfrac{3}{2}\\\\\\ -3A+C=-\dfrac{1}{2}\\\\\\ 3A-C=\dfrac{1}{2}$


Agora, resolva o sistema:

     $\left\{\begin{array}{rcl}A+C&\!\!\!=\!\!\!&\dfrac{1}{2}\\\\3A-C&\!\!\!=\!\!\!&\dfrac{1}{2} \end{array}\right.$


Somando as equações membro a membro, você obtém

     $A+C+3A-C=\dfrac{1}{2}+\dfrac{1}{2}\\\\\\ A+3A=1\\\\ 4A=1\\\\ A=\dfrac{1}{4}$


Para encontrar C, substituímos em uma das equações do sistema:

     $A+C=\dfrac{1}{2}\\\\\\ \dfrac{1}{4}+C=\dfrac{1}{2}\\\\\\ C=\dfrac{1}{2}-\dfrac{1}{4}\\\\\\ C=\dfrac{1}{4}$


Logo, a função a ser integrada já decomposta em frações parciais é

     $\dfrac{1}{(1-u)^2(1+u)^2}=\dfrac{\frac{1}{4}}{1-u}+\dfrac{\frac{1}{4}}{(1-u)^2}+\dfrac{\frac{1}{4}}{1+u}+\dfrac{\frac{1}{4}}{(1+u)^2}$


Portanto, a nossa integral fica

     $\displaystyle=\int\bigg[\frac{\frac{1}{4}}{1-u}+\frac{\frac{1}{4}}{(1-u)^2}+\frac{\frac{1}{4}}{1+u}+\frac{\frac{1}{4}}{(1+u)^2}\bigg]du\\\\\\ =\frac{1}{4}\int\frac{1}{1-u}\,du+\frac{1}{4}\int\frac{1}{(1-u)^2}\,du+\frac{1}{4}\int\frac{1}{1+u}\,du+\frac{1}{4}\int\frac{1}{(1+u)^2}\,du$

     $=-\dfrac{1}{4}\ln|1-u|-\dfrac{1}{4}\cdot \dfrac{(1-u)^{-2+1}}{-2+1}+\dfrac{1}{4}\ln|1+u|+\dfrac{1}{4}\cdot \dfrac{(1+u)^{-2+1}}{-2+1}+C$

     $=-\dfrac{1}{4}\ln|1-u|-\dfrac{1}{4}\cdot \dfrac{(1-u)^{-1}}{-1}+\dfrac{1}{4}\ln|1+u|+\dfrac{1}{4}\cdot \dfrac{(1+u)^{-1}}{-1}+C$

     $=-\dfrac{1}{4}\ln|1-u|+\dfrac{1}{4(1-u)}+\dfrac{1}{4}\ln|1+u|-\dfrac{1}{4(1+u)}+C$


Substitua de volta para a variável x:

     $=-\dfrac{1}{4}\ln|1-\mathrm{sen\,}x|+\dfrac{1}{4(1-\mathrm{sen\,}x)}+\dfrac{1}{4}\ln|1+\mathrm{sen\,}x|-\dfrac{1}{4(1+\mathrm{sen\,}x)}+C$

esta é a resposta.


Dúvidas? Comente.


Bons estudos! :-)