Question

alguém pode mostrar a resolução dessa integral?
$\int\limits {cosecx} \, dx$

Answer 1

Calcular a integral indefinida:

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


Faça a seguinte substituição:

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


e a integral fica

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


A integral acima pode ser resolvida decompondo o integrando em frações parciais:

     $f(u)=\dfrac{1}{(u-1)(u+1)}=\dfrac{A}{u-1}+\dfrac{B}{u+1}\\\\\\ \dfrac{1}{(u-1)(u+1)}=\dfrac{A(u+1)+B(u-1)}{(u-1)(u+1)}\\\\\\ \dfrac{0u+1}{(u-1)(u+1)}=\dfrac{(A+B)u+A-B}{(u-1)(u+1)}$


Por identidade polinomial dos numeradores, obtemos o sistema

     $\left\{\! \begin{array}{l} A+B=0\\\\ A-B=1 \end{array} \right.$


de onde obtemos

     $A=\dfrac{1}{2}\textsf{~~e~~}B=-\,\dfrac{1}{2}.$


Assim, a integral fica

     $\displaystyle=\int \left(\frac{\frac{1}{2}}{u-1}-\frac{\frac{1}{2}}{u+1}\right)du\\\\\\ =\frac{1}{2}\int \frac{1}{u-1}\,du-\frac{1}{2}\int\frac{1}{u+1}\,du\\\\\\ =\frac{1}{2}\ln\!|u-1|-\frac{1}{2}\ln\!|u+1|+C\\\\\\ =\frac{1}{2}\ln\!\left|\cos x-1\right|-\frac{1}{2}\ln\!\left|\cos x+1\right|+C$


Caso prefira pode parar por aqui. Mas podemos tentar simplificar a expressão para esta primitiva:

     $=\dfrac{1}{2}\cdot \big(\ln\!\left|\cos x-1\right|-\ln\!\left|\cos x+1\right|\big)+C\\\\\\ =\dfrac{1}{2}\ln\!\dfrac{\left|\cos x-1\right|}{\left|\cos x+1\right|}+C\\\\\\ =\dfrac{1}{2}\ln\left|\dfrac{\cos x-1}{\cos x+1}\right|+C\\\\\\ =\dfrac{1}{2}\ln\left|\dfrac{(\cos x-1)\cdot (\cos x-1)}{(\cos x+1)\cdot (\cos x-1)}\right|+C\\\\\\ =\dfrac{1}{2}\ln\left|\dfrac{(\cos x-1)^2}{\cos^2 x-1}\right|+C$

     $=\dfrac{1}{2}\ln\left|\dfrac{(\cos x-1)^2}{-\,\mathrm{sen^2\,}x}\right|+C\\\\\\ =\dfrac{1}{2}\ln\left|\dfrac{(\cos x-1)^2}{\mathrm{sen^2\,}x}\right|+C\\\\\\ =\dfrac{1}{2}\ln\!\left(\dfrac{\cos x-1}{\mathrm{sen\,}x}\right)^{\!2}+C\\\\\\ =\dfrac{1}{\diagup\!\!\!\! 2}\cdot \diagup\!\!\!\! 2\ln\!\left|\dfrac{\cos x-1}{\mathrm{sen\,}x}\right|+C\\\\\\ =\ln\!\left|\dfrac{\cos x}{\mathrm{sen\,}x}-\dfrac{1}{\mathrm{sen\,}x}\right|+C$

     $=\ln\!\left|\mathrm{cotg\,}x-\mathrm{cossec\,}x\right|+C\quad\longleftarrow\quad\textsf{resposta.}$


Bons estudos! :-)