Question

4) Resolva a integral e diga o método que está usando:

$\int\limits {tan^5~x} \, dx$

Answer 1

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador:  https://brainly.com.br/tarefa/10635668

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Calcular a integral indefinida:

     $\mathsf{\displaystyle\int\!tan^5\,x\,dx}$


Vamos manipular o integrando de forma a simplificar as potências de tangente.  Usaremos a identidade trigonométrica quando necessário:

     •   tan² x = sec² x − 1


     $\mathsf{\displaystyle\int\!tan^5\,x\,dx}\\\\\\ =\mathsf{\displaystyle\int\!tan^3\,x\cdot tan^2\,x\,dx}\\\\\\ =\mathsf{\displaystyle\int\!tan^3\,x\cdot (sec^2\,x-1)\,dx}\\\\\\ =\mathsf{\displaystyle\int\!(tan^3\,x\cdot sec^2\,x-tan^3\,x)\,dx}\\\\\\ =\mathsf{\displaystyle\int\! tan^3\,x\cdot sec^2\,x\,dx-\int\!tan^3\,x\,dx}\\\\\\ =\mathsf{\displaystyle\int\! tan^3\,x\cdot sec^2\,x\,dx-\int\!tan\,x\cdot tan^2\,x\,dx}$

     $=\mathsf{\displaystyle\int\! tan^3\,x\cdot sec^2\,x\,dx-\int\!tan\,x\cdot (sec^2\,x-1)\,dx}\\\\\\ =\mathsf{\displaystyle\int\! tan^3\,x\cdot sec^2\,x\,dx-\int\!(tan\,x\cdot sec^2\,x-tan\,x)\,dx}\\\\\\ =\mathsf{\displaystyle\int\! tan^3\,x\cdot sec^2\,x\,dx-\int\!tan\,x\cdot sec^2\,x\,dx+\int\!tan\,x\,dx\qquad\quad (i)}$


Para as duas primeiras integrais acima,  faça a seguinte substituição:

     $\mathsf{tan\,x=u\quad\Rightarrow \quad sec^2\,x\,dx=du}$


Para a terceira integral,  escreva tangente como o quociente de seno por cosseno.  A integral  (i)  fica

     $=\mathsf{\displaystyle\int\! u^3\,du-\int\!u\,du+\int\!\frac{sen\,x}{cos\,x}\,dx}\\\\\\ =\mathsf{\displaystyle\int\! u^3\,du-\int\!u\,du-\int\!\frac{1}{cos\,x}\cdot (-\,sen\,x)\,dx}$


Na terceira integral, faça a substituição

     $\mathsf{cos\,x=v\quad \Rightarrow\quad -\,sen\,x\,dx=dv}$


e fica

     $=\mathsf{\displaystyle\int\! u^3\,du-\int\!u\,du-\int\!\frac{1}{v}\,dv}\\\\\\ =\mathsf{\dfrac{u^{3+1}}{3+1}-\dfrac{u^{1+1}}{1+1}-\ell n|v|+C}\\\\\\ =\mathsf{\dfrac{u^4}{4}-\dfrac{u^2}{2}-\ell n|v|+C}$

     $=\mathsf{\dfrac{tan^4\,x}{4}-\dfrac{tan^2\,x}{2}-\ell n|cos\,x|+C}$   <————   esta é a resposta.


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int tan ^5\left(x\right)dx\\\\\\=\int \left(-1+sec ^2\left(x\right)\right)^2~tan \left(x\right)dx\\\\\\=\int \frac{\left(-1+u^2\right)^2}{u}du\\\\\\=\int \frac{1}{u}-2u+u^3du\\\\\\=\int \frac{1}{u}du-\int \:2udu+\int \:u^3du\\\\\\ln \left|sec \left(x\right)\right|-sec ^2\left(x\right)+\frac{sec ^4\left(x\right)}{4}\\\\\\\to \boxed{\sf ln \left|sec \left(x\right)\right|-sec ^2\left(x\right)+\frac{sec ^4\left(x\right)}{4}+C}$