Question

Qual a integral de sen(lnx^2)? preciso da resolução, obrigado!

Answer 1

$I=\displaystyle\int{\mathrm{sen}[\ln(x^{2})]\,dx}$


Método de integração por partes:

$\begin{array}{lcl} u=\mathrm{sen}[\ln(x^{2})]&~\Rightarrow~&du=\cos[\ln(x^{2})]\cdot \dfrac{1}{x^{2}}\cdot 2x\,dx\\ \\ dv=dx&~\Leftarrow~&v=x \end{array}\\ \\ \\ \\ \displaystyle\int{u\,dv}=uv-\int{v\,du}\\ \\ \\ \int{\mathrm{sen}[\ln(x^{2})]\,dx}=x\,\mathrm{sen}[\ln(x^{2})]-\int{x\cdot \cos[\ln(x^{2})]\cdot \dfrac{1}{x^{2}}\cdot 2x\,dx}\\ \\ \\ I=x\,\mathrm{sen}[\ln(x^{2})]-\underbrace{\int{2\cos[\ln(x^{2})]\,dx}}_{I_{1}}~~~~~~\mathbf{(i)}$


Calcularemos $I_{1}$ utilizando o método de integração por partes mais uma vez:

$I_{1}=\displaystyle\int{2\cos[\ln(x^{2})]\,dx}\\ \\ \\ \begin{array}{lcl} u=\cos[\ln(x^{2})]&~\Rightarrow~&du=-\mathrm{sen}[\ln(x^{2})]\cdot \dfrac{1}{x^{2}}\cdot 2x\,dx\\ \\ dv=2\,dx&~\Leftarrow~&v=2x \end{array}\\ \\ \\ \\ \int{u\,dv}=uv-\int{v\,du}\\ \\ \\ \int{2\cos[\ln(x^{2})]\,dx}=2x\cos[\ln(x^{2})]-\int{2x\cdot \left(-\mathrm{sen}[\ln(x^{2})]\cdot \dfrac{1}{x^{2}}\cdot 2x\,dx \right )}\\ \\ \\ I_{1}=2x\cos[\ln(x^{2})]+4\int{\mathrm{sen}[\ln(x^{2})\,dx}\\ \\ \\ I_{1}=2x\cos[\ln(x^{2})]+4I~~~~~~\mathbf{(ii)}$


Substituindo $\mathbf{(ii)}$ em $\mathbf{(i)},$ temos

$I=x\,\mathrm{sen}[\ln(x^{2})]-\left(2x\cos[\ln(x^{2})]+4I \right )\\ \\ I=x\,\mathrm{sen}[\ln(x^{2})]-2x\cos[\ln(x^{2})]-4I\\ \\ I+4I=x\,\mathrm{sen}[\ln(x^{2})]-2x\cos[\ln(x^{2})]\\ \\ 5I=x\,\mathrm{sen}[\ln(x^{2})]-2x\cos[\ln(x^{2})]\\ \\ I=\dfrac{1}{5}\cdot \left(x\,\mathrm{sen}[\ln(x^{2})]-2x\cos[\ln(x^{2})] \right )\\ \\ \\ I=\dfrac{1}{5}\,x\,\mathrm{sen}[\ln(x^{2})]-\dfrac{2}{5}\,x\cos[\ln(x^{2})]+C\\ \\ \\ \\ \Rightarrow~\boxed{ \begin{array}{c} \displaystyle\int{\mathrm{sen}[\ln(x^{2})]\,dx}=\dfrac{1}{5}\,x\,\mathrm{sen}[\ln(x^{2})]-\dfrac{2}{5}\,x\cos[\ln(x^{2})]+C \end{array} }$