Question

Derive a função q(x) = ftgx_e^-2x eθ cosθdθ.​

Answer 1

Desejamos calcular a derivada da função q(x), sendo a mesma dada da seguinte forma:

$\Large\displaystyle\text{ \begin{gathered} \tt q(x)= \int ^{\tan(x)}_{e^{-2x}}e^\theta \cos(\theta)d\theta \end{gathered} }$

Aplicando a integração por partes, temos que:

$\Large\displaystyle\begin{gathered} \tt u=\cos(\theta) \implies du=-\sin(\theta) d\theta \end{gathered}$

$\Large\displaystyle\begin{gathered} \tt dv=e^\theta d\theta \implies v=e^\theta\end{gathered}$

E com isso surge que:

$\Large\displaystyle\text{ \begin{gathered} \tt q(x)=e^\theta \cos(\theta)\Big|_{e^{-2x}}^{\tan(x)}+\int _{e^{-2x}}^{\tan(x)} e^\theta \sin(\theta)d\theta \end{gathered} }$

Aplicando novamente a integração por partes, ficamos da seguinte forma:

$\Large\displaystyle\begin{gathered} \tt u=\sin(\theta) \implies du=\cos(\theta) d\theta \end{gathered}$

$\Large\displaystyle\begin{gathered} \tt dv=e^\theta d\theta \implies v=e^\theta\end{gathered}$

Logo:

$\Large\displaystyle\text{ \begin{gathered} \tt q(x)=e^\theta \cos(\theta)\Big|_{e^{-2x}}^{\tan(x)}+\left[e^\theta\sin(\theta)\Big|_{e^{-2x}}^{\tan(x)}-\int _{e^{-2x}}^{\tan(x)}e^\theta \cos(\theta)d\theta\right] \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \tt q(x)=e^\theta \cos(\theta)\Big|_{e^{-2x}}^{\tan(x)}+e^\theta \sin(\theta)\Big|_{e^{-2x}}^{\tan(x)}-\int _{e^{-2x}}^{\tan(x)}e^\theta \cos(\theta)d\theta \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \tt 2q(x)=e^\theta \cos(\theta)\Big|_{e^{-2x}}^{\tan(x)}+e^\theta \sin(\theta)\Big|_{e^{-2x}}^{\tan(x)} \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \tt q(x)=\frac{e^\theta \cos(\theta)\Big|_{e^{-2x}}^{\tan(x)}+e^\theta \sin(\theta)\Big|_{e^{-2x}}^{\tan(x)}}{2} \end{gathered} }$

Aplicando o teorema fundamental do cálculo e derivando, temos por fim que a q'(x) é igual a:

$\Large\displaystyle\text{ \begin{gathered} \boxed{\tt q'(x)=e^{\tan \left(x\right)}\sec ^2\left(x\right)\cos \left(\tan \left(x\right)\right)+2e^{-2x+e^{-2x}}\cos \left(e^{-2x}\right)}\end{gathered} }$

Veja mais sobre:

• brainly.com.br/tarefa/48616510