Question

Determine as integrais:
b)∫$\frac{y}{2y+1}dx$
d)∫sen²2ФdФ

Answer 1

b) $\displaystyle\int{\dfrac{y}{2y+1}\,dy}$

$=\displaystyle\dfrac{1}{2}\int{\dfrac{2y}{2y+1}\,dy}\\ \\ \\ =\dfrac{1}{2}\int{\dfrac{2y+1-1}{2y+1}\,dy}\\\\\\ =\dfrac{1}{2}\int{\left(\dfrac{2y+1}{2y+1}-\dfrac{1}{2y+1} \right )\,dy}\\ \\ \\ =\dfrac{1}{2}\int{\left(1-\dfrac{1}{2y+1} \right )\,dy}\\ \\ \\ =\dfrac{1}{2}\int{\left(1-\dfrac{1}{2}\cdot \dfrac{2}{2y+1} \right )\,dy}\\ \\ \\ =\dfrac{1}{2}\int{dy}-\dfrac{1}{4}\int{\dfrac{2}{2y+1}\,dy}\\ \\ \\ =\dfrac{1}{2}\,y-\dfrac{1}{4}\int{\dfrac{2}{2y+1}\,dy}~~~~~~\mathbf{(i)}$


Fazendo a seguinte substituição:

$2y+1=u~~\Rightarrow~~2\,dy=du$


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

$=\displaystyle\dfrac{1}{2}\,y-\dfrac{1}{4}\int{\dfrac{1}{u}\,du}\\ \\ \\ =\dfrac{1}{2}\,y-\dfrac{1}{4}\cdot \mathrm{\ell n}\!\left|u\right|+C\\ \\ \\ =\dfrac{1}{2}\,y-\dfrac{1}{4}\,\mathrm{\ell n}\left|2y+1\right|+C$

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d) $\displaystyle\int{\mathrm{sen^{2}\,}2\phi\,d\phi}$

Usando a seguinte identidade trigonométrica:

$\mathrm{sen^{2}\,}\alpha=\dfrac{1}{2}-\dfrac{1}{2}\cos 2\alpha$


com $\alpha=2\phi,$ a integral fica

$=\displaystyle\int{\left(\dfrac{1}{2}-\dfrac{1}{2}\cos 4\phi \right )d\phi}\\ \\ \\ =\dfrac{1}{2}\int{d\phi}-\dfrac{1}{2}\int{\cos 4\phi\,d\phi}\\ \\ \\ =\dfrac{1}{2}\,\phi-\dfrac{1}{2}\cdot \dfrac{1}{4}\int{4\cos 4\phi\,d\phi}\\ \\ \\ =\dfrac{1}{2}\,\phi-\dfrac{1}{8}\int{4\cos 4\phi\,d\phi}\\ \\ \\ =\dfrac{1}{2}\,\phi-\dfrac{1}{8}\int{\cos 4\phi\cdot 4\,d\phi}~~~~~~\mathbf{(i)}$


Fazendo a seguinte substituição:

$4\phi=u~~\Rightarrow~~4\,d\phi=du$


Substituindo em $\mathbf{(i)},$ obtemos

$=\displaystyle\dfrac{1}{2}\,\phi-\dfrac{1}{8}\int{\cos u\,du}\\ \\ \\ =\dfrac{1}{2}\,\phi-\dfrac{1}{8}\cdot \mathrm{sen\,}u+C\\ \\ \\ =\dfrac{1}{2}\,\phi-\dfrac{1}{8}\,\mathrm{sen\,}4\phi+C$