Question

O estudo das ondas de dentes de serra em engenharia leva a integrais da forma

$\Large\displaystyle\text{ \begin{gathered}\int_{ \frac{-\pi}{\omega}}^{ \frac{\pi}{\omega}}\rm t\,sen(k\omega t)\,dt \end{gathered} }$

onde k é um inteiro e $\Large\omega$ é uma constante não nula. Calcule a integral.​

Answer 1

Resposta:   $\displaystyle\int_{-\pi/\omega}^{\pi/\omega}t\,\mathrm{sen}(k\omega t)\,dt=\left\{\begin{array}{ll}0~\mathrm{(zero)},&\mathrm{se~}k=0\\\\ (-1)^{k+1}\cdot \dfrac{2\pi}{k\omega^2},&\mathrm{se~}k\ne 0\end{array}\right..$

Explicação passo a passo:

Calcular a integral definida:

    $\displaystyle\int_{-\pi/\omega}^{\pi/\omega} t\,\mathrm{sen}(k\omega t)\,dt$

com k inteiro e ω ≠ 0.

• Se k = 0, a integral fica

    $=\displaystyle\int_{-\pi/\omega}^{\pi/\omega} t\,\mathrm{sen}(0\cdot \omega t)\,dt\\\\\\=\int_{-\pi/\omega}^{\pi/\omega} t\,\mathrm{sen}(0)\,dt\\\\\\=\int_{-\pi/\omega}^{\pi/\omega} 0\,dt=0.$

• Se k ≠ 0, vamos integrar por partes.

    $\begin{array}{lcl}u=t&\quad\Longrightarrow\quad&du=dt\\\\dv=\mathrm{sen}(k\omega t)\,dt&\quad\Longleftarrow\quad&v=-\,\dfrac{1}{k\omega}\cos(k\omega t)\end{array}$

    $\displaystyle\int_a^b u\,dv=uv\Big|_a^b-\int_a^b v\,du$

    $\begin{array}{lcr}\displaystyle\Longrightarrow \quad\int_{-\pi/\omega}^{\pi/\omega}t\,\mathrm{sen}(k\omega t)\,dt&\!\!=\!\!& t\cdot \left[-\,\dfrac{1}{k\omega}\cos(k\omega t)\right]_{-\pi/\omega}^{\pi/\omega}\\\\&&\displaystyle -\int_{-\pi/\omega}^{\pi/\omega}\left[-\,\frac{1}{k\omega}\cos(k\omega t)\right]\!dt\end{array}$

    $\displaystyle=\left.-\,\frac{t}{k\omega}\cos(k\omega t)\right|_{-\pi/\omega}^{\pi/\omega}+\frac{1}{k\omega}\int_{-\pi/\omega}^{\pi/\omega}\cos(k\omega t)\,dt\\\\\\=\left.-\,\frac{t}{k\omega}\cos(k\omega t)\right|_{-\pi/\omega}^{\pi/\omega}+\frac{1}{k\omega}\cdot \left.\frac{1}{k\omega}\,\mathrm{sen}(k\omega t)\right|_{-\pi/\omega}^{\pi/\omega}$

    $\displaystyle=\left.-\,\frac{t}{k\omega}\cos(k\omega t)\right|_{-\pi/\omega}^{\pi/\omega}+\left.\frac{1}{(k\omega)^2}\,\mathrm{sen}(k\omega t)\right|_{-\pi/\omega}^{\pi/\omega}$

    $\begin{array}{lr}=\!\!\!&\left[-\dfrac{~\frac{\pi}{\omega}~}{k\omega}\cos\!\left(k\omega\cdot \dfrac{\pi}{\omega}\right)+\dfrac{~-\frac{\pi}{\omega}~}{k\omega}\cos\!\left(k\omega \cdot -\dfrac{\pi}{\omega}\right)\right]\\\\&+\left[\dfrac{1}{(k\omega)^2}\,\mathrm{sen}\!\left(k\omega\cdot \dfrac{\pi}{\omega}\right)-\dfrac{1}{(k\omega)^2}\,\mathrm{sen}\!\left(k\omega\cdot -\dfrac{\pi}{\omega}\right)\right]\end{array}$

    $=\left[-\dfrac{\pi}{k\omega^2}\cos(k\pi)-\dfrac{\pi}{k\omega^2}\cos(-k\pi)\right]+\underbrace{\left[\dfrac{1}{(k\omega)^2}\,\mathrm{sen}(k\pi)-\dfrac{1}{(k\omega)^2}\,\mathrm{sen}(-k\pi)\right]}_{=0\mathrm{~(zero),~pois}~sen(\pm\,k\pi)=0}$

    $=\left[-\dfrac{\pi}{k\omega^2}\cos(k\pi)-\dfrac{\pi}{k\omega^2}\cos(-k\pi)\right]+0\\\\\\=-\dfrac{\pi}{k\omega^2}\cos(k\pi)-\dfrac{\pi}{k\omega^2}\cos(-k\pi)$

Como o cosseno é uma função par, temos cos(− kπ) = cos(kπ). Então, o resultado fica

    $=-\dfrac{\pi}{k\omega^2}\cos(k\pi)-\dfrac{\pi}{k\omega^2}\cos(k\pi)\\\\\\=-\,\dfrac{2\pi}{k\omega^2}\cos(k\pi)$

    $=\left\{\begin{array}{ll}-\,\dfrac{2\pi}{k\omega^2}\cdot 1,&\mathrm{se~}k\mathrm{~for~par}\\\\-\,\dfrac{2\pi}{k\omega^2}\cdot (-1),&\mathrm{se~}k\mathrm{~for~\'impar}\end{array}\right.$

    $=\left\{\begin{array}{rl}-\,\dfrac{2\pi}{k\omega^2},&\mathrm{se~}k\mathrm{~for~par}\\\\\dfrac{2\pi}{k\omega^2},&\mathrm{se~}k\mathrm{~for~\'impar}\end{array}\right.$

    $=(-1)^{k+1}\cdot \dfrac{2\pi}{k\omega^2}\quad\longleftarrow\quad \mathsf{resposta~para~}k\ne 0.$

Dúvidas? Comente.

Bons estudos.