Question

Usando as propriedades e as técnicas de integração, bem como sabendo as integrais imediatas, calcule a integral indefinida abaixo:

$\mathsf{\displaystyle \int \dfrac{dx}{(1-sen\ x)^2}}$

Answer 1

Calcular a integral indefinida:

$\displaystyle\int\frac{1}{(1-sen\,x)^2}\,dx\\\\\\ =\int\frac{1}{\left(sen\,\frac{\pi}{2}-sen\,x\right)^2}\,dx$

Aplicamos agora algumas identidades trigonométricas:

$\left\{\!\begin{array}{l}sen\,a-sen\,b=2\,sen\!\left(\dfrac{a-b}{2}\right)cos\!\left(\dfrac{a+b}{2}\right)\\\\ sen(-a)=-\,sen(a)\\\\ cos\,a=sen\!\left(\dfrac{\pi}{2}-a\right) \end{array} \right.$

e a integral fica

$\displaystyle\int\frac{1}{\left[2\,sen\!\left(\dfrac{\frac{\pi}{2}-x}{2}\right)cos\!\left(\dfrac{\frac{\pi}{2}+x}{2}\right)\right]^{\!2}}\,dx\\\\\\ =\int\frac{1}{\left[2\,sen\!\left(\frac{\pi}{4}-\frac{x}{2}\right)cos\!\left(\frac{\pi}{4}+\frac{x}{2}\right)\right]^{\!2}}\,dx\\\\\\ =\int\frac{1}{\left[2\,sen\!\left(\frac{\pi}{4}-\frac{x}{2}\right)sen\!\left(\frac{\pi}{2}-\big(\frac{\pi}{4}+\frac{x}{2}\big)\right)\right]^{\!2}}\,dx\\\\\\ =\int\frac{1}{\left[2\,sen\!\left(\frac{\pi}{4}-\frac{x}{2}\right)sen\!\left(\frac{\pi}{4}-\frac{x}{2}\right)\right]^{\!2}}\,dx\\\\\\ =\int\frac{1}{\left[2\,sen^2\!\left(\frac{\pi}{4}-\frac{x}{2}\right)\right]^{\!2}}\,dx\\\\\\ =\int\frac{1}{\left[2\,sen^2\!\left(\frac{x}{2}-\frac{\pi}{4}\right)\right]^{\!2}}\,dx$

$\displaystyle=\int\frac{1}{4\,sen^4\!\left(\frac{x}{2}-\frac{\pi}{4}\right)}\,dx\\\\\\ =\frac{1}{4}\int cossec^4\!\left(\frac{x}{2}-\frac{\pi}{4}\right)\,dx\\\\\\ =\frac{1}{4}\int cossec^4\!\left(\frac{x}{2}-\frac{\pi}{4}\right)\cdot 2\cdot \frac{1}{2}\,dx\\\\\\ =\frac{1}{2}\int cossec^4\!\left(\frac{x}{2}-\frac{\pi}{4}\right)\cdot \frac{1}{2}\,dx$

Agora faça uma substituição:

$\dfrac{x}{2}-\dfrac{\pi}{4}=u\quad\Rightarrow\quad \dfrac{1}{2}\,dx=du$

e a integral fica

$\displaystyle=\frac{1}{2}\int cossec^4\,u\,du\\\\\\ =\frac{1}{2}\int cossec^2\,u\cdot cossec^2\,u\,du\\\\\\ =\frac{1}{2}\int (1+cotg^2\,u)\cdot cossec^2\,u\,du\\\\\\ =-\,\frac{1}{2}\int (1+cotg^2\,u)\cdot (-\,cossec^2\,u)\,du$

Faça outra substituição:

$cotg\,u=v\quad\Rightarrow\quad -\,cossec^2\,u\,du=dv$

e a integral fica

$\displaystyle=-\,\frac{1}{2}\int (1+v^2)\,dv\\\\\\ =-\,\frac{1}{2}\cdot \left(v+\frac{v^3}{3}\right)+C\\\\\\ =-\,\frac{1}{2}\,v-\frac{1}{6}\,v^3+C$

$=-\,\dfrac{1}{2}\,cotg\,u-\,\dfrac{1}{6}\,cotg^3\,u+C\\\\\\ =-\,\dfrac{1}{2}\,cotg\!\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)-\,\dfrac{1}{6}\,cotg^3\!\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)+C$

esta é a resposta.

Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \frac{dx}{\left(1-sin \left(x\right)\right)^2}\\\\\\=\int \frac{2\left(u^2+1\right)}{\left(1+u^2-2u\right)^2}du\\\\\\{Remova\:a\:constante}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\\\=2\cdot \int \frac{u^2+1}{\left(1+u^2-2u\right)^2}du\\\\\\=2\cdot \int \frac{1}{\left(u-1\right)^2}+\frac{2}{\left(u-1\right)^3}+\frac{2}{\left(u-1\right)^4}du$

$\sf \displaystyle =2\left(\int \frac{1}{\left(u-1\right)^2}du+\int \frac{2}{\left(u-1\right)^3}du+\int \frac{2}{\left(u-1\right)^4}du\right)\\\\\\=2\left(-\frac{1}{u-1}-\frac{1}{\left(u-1\right)^2}-\frac{2}{3\left(u-1\right)^3}\right)\\\\\\{Substitua\:na\:equação}\:u=tan \left(\frac{x}{2}\right)\\\\\\=2\left(-\frac{1}{tan \left(\frac{x}{2}\right)-1}-\frac{1}{\left(tan \left(\frac{x}{2}\right)-1\right)^2}-\frac{2}{3\left(tan \left(\frac{x}{2}\right)-1\right)^3}\right)$

$\to \boxed{\sf =2\left(-\frac{1}{tan \left(\frac{x}{2}\right)-1}-\frac{1}{\left(tan \left(\frac{x}{2}\right)-1\right)^2}-\frac{2}{3\left(tan \left(\frac{x}{2}\right)-1\right)^3}\right)+C}$