Question

Determine as integrais:
b)∫$\frac{y}{ \sqrt{2y+1} } dy$
d)∫sen³2ФdФ

Answer 1

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


Fazendo a seguinte substituição:

$2y+1=u~~\Rightarrow~~\left\{ \begin{array}{l} 2\,dy=du~~\Rightarrow~~dy=\dfrac{1}{2}\,du\\ \\ y=\dfrac{1}{2}\,(u-1) \end{array} \right.$


Substituindo, a integral fica

$=\displaystyle\int{\dfrac{\frac{1}{2}(u-1)}{\sqrt{u}}\cdot \dfrac{1}{2}\,du}\\ \\ \\ =\dfrac{1}{4}\int{\dfrac{u-1}{\sqrt{u}}\,du}\\ \\ \\ =\dfrac{1}{4}\int{\left(\dfrac{u}{\sqrt{u}}-\dfrac{1}{\sqrt{u}} \right )du}\\ \\ \\ =\dfrac{1}{4}\int{\left(\sqrt{u}-\dfrac{1}{\sqrt{u}} \right )du}\\ \\ \\ =\dfrac{1}{4}\int{\left(u^{1/2}-\dfrac{1}{u^{1/2}} \right )du}\\ \\ \\ =\dfrac{1}{4}\int{\left(u^{1/2}-u^{-1/2} \right )du}~~~~~~\mathbf{(i)}$


Regra para integral de potência:

$=\displaystyle\int{u^{n}\,du}=\dfrac{u^{n+1}}{n+1}\,,~~~\text{com } n\ne -1.$


Aplicando a regra da integral de potência em $\mathbf{(i)},$ a integral fica

$=\dfrac{1}{4}\cdot \left(\dfrac{u^{(1/2)+1}}{\frac{1}{2}+1}-\dfrac{u^{-(1/2)+1}}{-\frac{1}{2}+1} \right )+C\\ \\ \\ =\dfrac{1}{4}\cdot \left(\dfrac{u^{3/2}}{\frac{3}{2}}-\dfrac{u^{1/2}}{\frac{1}{2}} \right )+C\\ \\ \\ =\dfrac{1}{4}\cdot \left(\dfrac{2}{3}\,u^{3/2}-2\,u^{1/2} \right )+C\\ \\ \\ =\dfrac{1}{6}\,u^{3/2}-\dfrac{1}{2}\,u^{1/2}+C\\ \\ \\ =\dfrac{1}{6}\,(2y+1)^{3/2}-\dfrac{1}{2}\,(2y+1)^{1/2}+C$

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d) $\displaystyle\int{\mathrm{sen^{3}\,}2\phi\,d\phi}$

$=\displaystyle\int{\mathrm{sen^{2}\,}2\phi\cdot \mathrm{sen\,}2\phi\,d\phi}\\ \\ \\ =\int{(1-\cos^{2}2\phi)\cdot \mathrm{sen\,}2\phi\,d\phi}\\ \\ \\ =-\dfrac{1}{2}\int{(-2)\cdot (1-\cos^{2}2\phi)\cdot \mathrm{sen\,}2\phi\,d\phi}\\ \\ \\ =-\dfrac{1}{2}\int{(1-\cos^{2}2\phi)\cdot (-2)\,\mathrm{sen\,}2\phi\,d\phi}~~~~\mathbf{(i)}$


Fazendo a seguinte substituição:

$\cos 2\phi=u~\Rightarrow~-\!2\,\mathrm{sen\,}2\phi\,d\phi=du$


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

$=\displaystyle-\dfrac{1}{2}\int{(1-u^{2})\,du}\\ \\ \\ =-\dfrac{1}{2}\cdot \left(u-\dfrac{u^{3}}{3} \right )+C\\ \\ \\ =-\dfrac{u}{2}+\dfrac{u^{3}}{6}+C\\ \\ \\ =-\dfrac{\cos 2\phi}{2}+\dfrac{\cos^{3} 2\phi}{6}+C$