Question

∫sen 2x/raiz₅ de 1+3 cos (2x) dx.

Answer 1

$I=\displaystyle\int{\dfrac{\mathrm{sen\,}2x}{\,^{5}\!\!\!\sqrt{1+3\cos 2x}}\,dx}\\ \\ \\ =-\dfrac{1}{6}\cdot \int{\dfrac{(-6)\,\mathrm{sen\,}2x}{\,^{5}\!\!\!\sqrt{1+3\cos 2x}}\,dx}~~~~\mathbf{(i)}$


Vamos fazer a seguinte substituição:

$1+3\cos 2x=u~\Rightarrow~-6\,\mathrm{sen\,}2x\,dx=du$


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

$=-\dfrac{1}{6}\cdot \displaystyle\int{\dfrac{1}{\,^{5}\!\!\!\sqrt{u}}\,du}\\ \\ \\ =-\dfrac{1}{6}\cdot \int{\dfrac{1}{u^{1/5}}\,du}\\ \\ \\ =-\dfrac{1}{6}\cdot \int{u^{-1/5}\,du}~~~~\mathbf{(ii)}$


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 $\mathbf{(ii)},$ obtemos

$=-\dfrac{1}{6}\cdot \dfrac{u^{(-1/5)+1}}{-\frac{1}{5}+1}+C\\ \\ \\ =-\dfrac{1}{6}\cdot \dfrac{u^{4/5}}{\frac{4}{5}}+C\\ \\ \\ =-\dfrac{1}{6}\cdot \dfrac{5}{4}\,u^{4/5}+C\\ \\ \\ =-\dfrac{5}{24}\,u^{4/5}+C\\ \\ \\ =-\dfrac{5}{24}\,(1+3\cos 2x)^{4/5}+C\\ \\ \\ =-\dfrac{5}{24}\;^{5}\!\!\!\!\sqrt{(1+3\cos 2x)^{4}}+C$


Portanto,

$\boxed{\begin{array}{c} \displaystyle\int{\dfrac{\mathrm{sen\,}2x}{\,^{5}\!\!\!\sqrt{1+3\cos 2x}}\,dx}=-\dfrac{5}{24}\;^{5}\!\!\!\!\sqrt{(1+3\cos 2x)^{4}}+C \end{array}}$