Calcule a integral indefinida:.... Pergunta de ideia deLukyo

November 2019 1 141 Report

Calcule a integral indefinida:

$\displaystyle\int\frac{\sin 2x}{\cos 2x+\cos 8x}\,dx$

Lista de comentários

superaks

Verified answer

Olá Lukyo.

Identiddades utilizadas:

$\mathsf{\star~~\boxed{\boxed{\mathsf{sen(\alpha\pm\beta)=sen(\alpha)cos(\beta)\pm sen(\beta)cos(\alpha)}}}}\\\\\\\star~~\boxed{\boxed{\mathsf{cos(\alpha\pm\beta)=cos(\alpha)cos(\beta)\mp sen(\alpha)sen(\beta)}}}$

___________________

Organizando a integral.

$\mathsf{\displaystyle\int \dfrac{sen2x}{cos(2x)+cos(8x)}~dx}$

Vamos trabalhar agora somente com o denominador.

Faça:

$\begin{cases}\mathsf{\alpha+\beta=8x}\\\\\mathsf{\alpha-\beta=2x}\end{cases}\mathsf{~~\Rightarrow~~\alpha=5x~~e~~\beta=3x}$

Usando a identidade do cosseno, temos:

$\mathsf{cos(2x)+cos(8x)=cos(5x-3x)+cos(5x+3x)}\\\\\\\begin{cases}\mathsf{\cos(5x-3x)=cos(5x)cos(3x)+sen(5x)sen(3x)}\\+\\\mathsf{cos(5x+3x)=cos(5x)cos(3x)-sen(5x)sen(3x)}\\=\\\mathsf{cos(2x)+cos(5x)=2cos(5x)cos(3x)}\end{cases}}$

Substituindo a igualdade acima no denominador do integrando.

$\mathsf{\displaystyle\int \dfrac{sen2x}{cos(2x)+cos(8x)}~dx=\int\dfrac{sen2x}{2cos(5x)cos(3x)}~dx}\\\\\\\mathsf{\displaystyle\int\dfrac{sen2x}{cos(2x)+cos(5x)}~dx=\dfrac{1}{2}\cdot\int\displaystyle\dfrac{sen2x}{cos(5x)cos(3x)}~dx}$

Note que 2x = 5x - 3x.

Usando a identidade do seno, temos:

$\mathsf{sen(5x-3x)=sen(5x)cos(3x)-sen(3x)cos(5x)}$

Substitindo a igualdade acima no numerador do integrando.

$\mathsf{\displaystyle\int\dfrac{sen(2x)}{cos(2x)+cos(8x)}~dx=\dfrac{1}{2}\cdot\int\Big[\dfrac{sen(5x)cos(3x)-sen(3x)cos(5x)}{cos(5x)cos(3x)}\Big]~dx}\\\\\\\mathsf{\displaystyle\int\dfrac{sen(2x)}{cos(2x)+cos(8x)}~dx=\dfrac{1}{2}\cdot\int\Big[\dfrac{sen(5x)cos(3x)}{cos(5x)cos(3x)}-\dfrac{sen(3x)cos(5x)}{cos(5x)cos(3x)}\Big]~dx}\\\\\\\mathsf{\displaystyle\int\dfrac{sen(2x)}{cos(2x)+cos(8x)}~dx=\dfrac{1}{2}\cdot\int\Big[\dfrac{sen(5x)}{cos(5x)}-\dfrac{sen(3x)}{cos(3x)}\Big]~dx}$

$\mathsf{\displaystyle\int\dfrac{sen(2x)}{cos(2x)+cos(8x)}~dx=\dfrac{1}{2}\cdot\int [tg(5x)-tg(3x)]~dx}$

Faça:

$\mathsf{u = 5x~~~~~~e~~~~~v = 3x}\\\\\mathsf{ du = 5 dx~~~ e~~~~~~ dv = 3dx}$

$\mathsf{\displaystyle\int\dfrac{sen(2x)}{cos(2x)+cos(8x)}~dx=\dfrac{1}{2}\cdot\Big[\int\dfrac{tg(u)}{5}~du-\int\dfrac{tg(v)}{3}~dv\Big]}\\\\\\\mathsf{\displaystyle\int\dfrac{sen(2x)}{cos(2x)+cos(5x)}~dx=-\dfrac{\ell n|cos(u)|}{10}+\dfrac{\ell n|cos(v)|}{6}+C}$

Substituindo u e v.

$\mathsf{\displaystyle\int\dfrac{sen(2x)}{cos(2x)+cos(8x)}~dx=\dfrac{\ell n|cos(3x)|}{6}-\dfrac{\ell n|cos(5x)|}{10}+C}$

Dúvidas? Comente.

2 votes Thanks 4

Lukyo Obrigado! :-)

superaks Disponha !!