Question

Calculando a antiderivada:

$\large \displaystyle\int \dfrac{\sf sen \Bigg(\dfrac{x-1}{3}\Bigg) }{ \sf cos^2 \Bigg(\dfrac{x-1}{3} \Bigg)} dx$

Qual resultado obtemos?

Answer 1

• O resultado dessa integral é 3sec(x+1/3) + k.

Para resolver sua questão, devemos aplicar o método da substituição simples. Chamando assim:

$\large\displaystyle\begin{aligned} u = cos\left( \frac{x-1}{3}\right) \end{aligned}$

$\large\displaystyle\begin{aligned} du = -senx\left( \frac{x-1}{3} \right) \cdot \frac{1}{3}\ dx\end{aligned}$

$\large\displaystyle\begin{aligned} -du = senx\left( \frac{x-1}{3} \right) \cdot \frac{1}{3}\ dx\end{aligned}$

$\large\displaystyle\begin{aligned} -3du = senx\left( \frac{x-1}{3} \right) dx\end{aligned}$

Substituindo na sua questão, temos que:

$\large\displaystyle\text{ \begin{aligned} \int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = \int \frac{-3du}{u^2}\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = -3\cdot \int \frac{du}{u^2}\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = -3\cdot \int \frac{1}{u^2}du\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = -3\cdot \int u^{-2} du\end{aligned} }$

• Agora, aplique a seguinte propriedade de integração:

$\large\displaystyle\text{ \begin{aligned} \int x^ndx = \frac{x^{n+1}}{n+1} \ ,\ \underline{\underline{n\neq 1}}\end{aligned} }$

Continuando ...

$\large\displaystyle\text{ \begin{aligned} \int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = -3\cdot \int u^{-2} du\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = -3\cdot \frac{u^{-2+1} }{-2+1} + k\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = -3\cdot \frac{u^{-1} }{-1} + k\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = 3\cdot \frac{1}{u} + k\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = 3\cdot \frac{1}{cos\left( \frac{x-1}{3}\right)} + k\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \therefore\boxed{\boxed{\green{\int \frac{sen\left(\frac{x-1}{3}\right)}{cos^2 \left( \frac{x-1}{3}\right)}\ dx = 3sec\left( \frac{x-1}{3}\right) + k}}}\end{aligned} }$

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Integrais indefinidas.

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Answer 2

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{sen\bigg(\dfrac{x-1}{3}\bigg)}{cos^2\bigg(\dfrac{x-1}{3}\bigg)}~dx\\\underline{\rm Vamos ~multiplicar~e~dividir~por~3:}\\\displaystyle\sf3\int\dfrac{sen\bigg(\dfrac{x-3}{3}\bigg)}{cos^2\bigg(\dfrac{x-3}{3}\bigg)}\cdot\dfrac{1}{3}~dx\\\underline{\rm escrevamos~a~integral~de~outra~maneira,observe!}\\\displaystyle\sf3\int\dfrac{1}{cos\bigg(\dfrac{x-3}{3}\bigg)}\cdot\dfrac{sen\bigg(\dfrac{x-3}{3}\bigg)}{cos\bigg(\dfrac{x-3}{3}\bigg)}\cdot\dfrac{1}{3}~dx\end{array}}$

$\boxed{\begin{array}{l}\sf mas~\dfrac{1}{cos\bigg(\dfrac{x-3}{3}\bigg)}=sec\bigg(\dfrac{x-3}{3}\bigg)\\\\\sf\dfrac{sen\bigg(\dfrac{x-3}{3}\bigg)}{cos\bigg(\dfrac{x-3}{3}\bigg)}=tg\bigg(\dfrac{x-3}{3}\bigg)\\\\\sf substituindo~na~integral~temos:\\\displaystyle\sf3\int sec\bigg(\dfrac{x-3}{3}\bigg)\cdot tg\bigg(\dfrac{x-3}{3}\bigg)\cdot\dfrac{1}{3}~dx\\\underline{\boldsymbol{fac_{\!\!,}a}}\\\rm t=\dfrac{x-3}{3}\implies dt=\dfrac{1}{3}~dx\\\displaystyle\sf3\int sec(t)\cdot tg(t)~dt=3~sec(t)+k\\\sf substituindo~temos:\\\displaystyle\sf\int \dfrac{sen\bigg(\dfrac{x-3}{3}\bigg)}{cos^2\bigg(\dfrac{x-3}{3}\bigg)}~dx=3sec\bigg(\dfrac{x-3}{3}\bigg)+k\end{array}}$