Question

Calcule a seguinte integral definida de ∫sen(2x)dx no intervalo de pi/8 a 0

Answer 1

$\displaystyle\int_{0}^{\pi/8}\mathrm{sen}(2x)\,dx\\\\\\ =\int_{0}^{\pi/8}\dfrac{1}{2}\cdot 2\,\mathrm{sen}(2x)\,dx\\\\\\ =\dfrac{1}{2}\int_{0}^{\pi/8}\mathrm{sen}(2x)\cdot 2\,dx~~~~~~\mathbf{(i)}$


Mudança de variável:

$2x=u~~\Rightarrow~~2\,dx=du$


Mudando os extremos de integração:

$\text{Quando }x=0~~\Rightarrow~~u=0\\\\ \text{Quando }x=\dfrac{\pi}{8}~~\Rightarrow~~u=\dfrac{\pi}{4}$


Substituindo em $\mathbf{(i)},$ a integral fica

$=\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\mathrm{sen\,}u\,du\\\\\\ =\dfrac{1}{2}\cdot (-\cos u)\big|_{0}^{\pi/4}\\\\\\ =\dfrac{1}{2}\cdot \left[-\cos\dfrac{\pi}{4}+\cos 0 \right ]\\\\\\ =\dfrac{1}{2}\cdot \left(-\dfrac{\sqrt{2}}{2}+1 \right )\\\\\\=\dfrac{1}{2}\cdot \left(-\dfrac{\sqrt{2}}{2}+\dfrac{2}{2} \right )\\\\\\ =\dfrac{2-\sqrt{2}}{4}$


Bons estudos! :-)