Question

INTEGRAL
$\displaystyle\int_{-a}^{L-a} \frac{\lambda b}{4 \pi \epsilon_{o}(x^{2}+b^{2})^\frac{3}{2}}dx$

L = Comprimento
λ = Densidade de Carga por Unidade de Comprimento
ε = Permissividade no Vácuo

Answer 1

Resolução da questão, veja:

Resolver a integral:

$\mathsf{\displaystyle\int\limits_{-a}^{L-a} \dfrac{\lambda b}{4 \pi \epsilon_{o}(x^{2}+b^{2})^\frac{3}{2}}~dx}}$

Vamos puxar as constantes para fora da função e integrar o que resta, veja:

$\mathsf{\displaystyle\int\limits_{-a}^{L-a} \dfrac{\lambda b}{4 \pi \epsilon_{o}(x^{2}+b^{2})^\frac{3}{2}}~dx}}=\mathsf{\dfrac{\lambda b}{4\pi\epsilon_{o}}~\mathsf{\displaystyle\int\limits_{-a}^{L-a}~\dfrac{1}{(x^{2}+b^{2})^{\frac{3}{2}}}}~\mathsf{dx}}= \mathsf{\dfrac{\lambda b}{4\pi\epsilon_{o}}~\mathsf{\displaystyle\int\limits_{-a}^{L-a}~\dfrac{1}{\left[b^{2}~\left(\dfrac{x^{2}}{b^{2}}+1\right)\right]^{\frac{3}{2}}}}}~\mathsf{dx}}\\\\\\\\ \mathsf{\dfrac{\lambda b}{4\pi\epsilon_{o}}~\mathsf{\displaystyle\int\limits_{-a}^{L-a}~\dfrac{1}{b^{3}~\left[\left(\bigg(\dfrac{x}{b}\bigg)^{2}+1\right)\right]^{\frac{3}{2}}}}}~\mathsf{dx}}$

Façamos as seguintes substituições:

u = x/b = du = dx/b;

b • du = dx

Temos que:

$\mathsf{\dfrac{\lambda b}{4\pi\epsilon_{o}}~\mathsf{\displaystyle\int~\dfrac{1}{b^{3}~\left[\left(\bigg(\dfrac{x}{b}\bigg)^{2}+1\right)\right]^{\frac{3}{2}}}}}~\mathsf{dx}}=\mathsf{\displaystyle\int~\dfrac{1}{b^{2}~[u^{2}+1]^{\frac{3}{2}}}~\mathsf{du}}}$

Façamos mais algumas substituições:

u = tg y => du = sec²y dy;

y = arctan u;

tg²y + 1 = sec²y

Vejamos:

$\mathsf{\displaystyle\int~\dfrac{1}{b^{2}~[u^{2}+1]^{\frac{3}{2}}}~\mathsf{du}}}=\mathsf{\displaystyle\int~\dfrac{1}{b^{2}~[tg^{2}y+1]^{\frac{3}{2}}}~\cdot~\mathsf{sec^{2}y~dy}}}}\\\\\\\\ \mathsf{\displaystyle\int~\dfrac{1}{b^{2}~[sec^{2}y]^{\frac{3}{2}}}~\cdot~\mathsf{sec^{2}y~dy}}}}=\mathsf{\displaystyle\int~\dfrac{1}{b^{2}~sec^{2}y}}}~\mathsf{dy}}}$

Retirando as constantes teremos:

$\mathsf{\mathsf{\displaystyle\int~\dfrac{1}{b^{2}~sec^{2}y}}}~\mathsf{dy}}}=\mathsf{\dfrac{1}{b^{2}}~\displaystyle\int~\dfrac{1}{sec~y}}~\mathsf{dy}}}=\mathsf{\dfrac{1}{b^{2}}~\cdot~\displaystyle\int~cos~y~dy}}}=\mathsf{\dfrac{sen~y}{b^{2}}}\\\\\\\\ \mathsf{\dfrac{sen~(arctan~u)}{b^{2}}}$

Fazendo k = arctan u, deste modo, $\mathsf{sen~k=\dfrac{u}{\sqrt{u^{2}+1}}}$, utilizando a relação $\mathsf{tan~k=\dfrac{sen~k}{cos~k}}$ e sen²k + cos²k = 1, com:

tan k = u;

$\mathsf{sen~k=\dfrac{u}{\sqrt{u^{2}+1}}}}$

Vejamos:

$\mathsf{\dfrac{sen~(arctan~u)}{b^{2}}}=\mathsf{\dfrac{u}{b^{2}~\sqrt{u^{2}+1}}}$

Como u = x/b, a expressão acima se converte em:

$\mathsf{\dfrac{u}{b^{2}~\sqrt{u^{2}+1}}}=\mathsf{\dfrac{\frac{x}{b}}{b^{2}~\sqrt{u^{2}+1}}}}=\mathsf{\dfrac{x}{b^{2}~\sqrt{u^{2}+1}}}$

Deste modo:

$\mathsf{\dfrac{\lambda b}{4\pi\epsilon_{o}}~\mathsf{\displaystyle\int~\dfrac{1}{b^{3}~\left[\left(\bigg(\dfrac{x}{b}\bigg)^{2}+1\right)\right]^{\frac{3}{2}}}}}~\mathsf{dx}}=\mathsf{\dfrac{\lambda b}{4\pi\epsilon_{o}}~\cdot~\left[\dfrac{x}{b^{2}~\sqrt{x^{2}+b^{2}\right]}}}\bigg|_{-a}^{L-a}}}\\\\\\\\ \mathsf{\dfrac{\lambda b}{4\pi\epsilon_{o}}}~\mathsf{\left(\dfrac{L-a}{b^{2}~\sqrt{(L-a)^{2}+b^{2}}}+\mathsf{\dfrac{a}{b^{2}~\sqrt{a^{2}+b^{2}}\right)}}}}\\\\\\\\ \mathsf{\dfrac{\lambda b}{4\pi\epsilon_{o}~\cdot~b}}~\mathsf{\left(\dfrac{L-a}{\sqrt{(L-a)^{2}+b^{2}}}+\mathsf{\dfrac{a}{b^{2}~\sqrt{a^{2}+b^{2}}\right)}}}}}}$

Portanto, podemos dizer que:

$\mathsf{\displaystyle\int\limits_{-a}^{L-a}~\dfrac{\lambda b}{4 \pi \epsilon_{o}(x^{2}+b^{2})^\frac{3}{2}}~dx}}= \mathsf{\dfrac{\lambda b}{4\pi\epsilon_{o}~\cdot~b}}~\mathsf{\left(\dfrac{L-a}{\sqrt{(L-a)^{2}+b^{2}}}+\mathsf{\dfrac{a}{b^{2}~\sqrt{a^{2}+b^{2}}\right)}}}}}}$

Espero que te ajude. (^.^)