Question

calcule o integral:
$\displaystyle\int\sf{\cos\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{3}\right)dx}$​

Answer 1

$3 \sin( \frac{x}{6} ) + \frac{3}{5} \sin( \frac{5x}{6} ) + c$

Answer 2

$\boxed{\begin{array}{l}\underline{\rm Identidade~de~Prostaf\acute erese}\\\sf cos(\alpha)\cdot cos(\theta)=\dfrac{1}{2}\cdot[cos(\alpha+\theta)+cos(\alpha-\theta)]\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int cos\bigg(\dfrac{x}{2}\bigg) cos\bigg(\dfrac{x}{3}\bigg)~dx=\int\dfrac{1}{2}\bigg[cos\bigg(\dfrac{x}{2}+\dfrac{x}{3}\bigg)+cos\bigg(\dfrac{x}{2}-\dfrac{x}{3}\bigg)\bigg]dx\\\\\displaystyle\sf\dfrac{1}{2}\int \bigg[cos\bigg(\dfrac{5x}{6}\bigg)+cos\bigg(\dfrac{x}{6}\bigg)\bigg]dx=\dfrac{1}{2}\int cos\bigg(\dfrac{5x}{6}\bigg)dx+\dfrac{1}{2}\int cos\bigg(\dfrac{x}{6} \bigg)dx\\\displaystyle\sf\int cos(at)~dt=\dfrac{1}{a} sen(at)+k\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int cos\bigg(\dfrac{5x}{6}\bigg)dx=\dfrac{1}{\frac{5}{6}} sen\bigg(\dfrac{5x}{6}\bigg)+k=\dfrac{6}{5} sen\bigg(\dfrac{5x}{6}\bigg)+k\\\\\displaystyle\sf\int cos\bigg(\dfrac{x}{6}\bigg)dx=\dfrac{1}{\frac{1}{6}}sen\bigg(\dfrac{x}{6}\bigg)+k=6 sen\bigg(\dfrac{x}{6}\bigg)+k\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf \int cos\bigg(\dfrac{x}{2}\bigg)~cos\bigg(\dfrac{x}{3}\bigg)dx=\dfrac{1}{\backslash\!\!\!2}\cdot\dfrac{\backslash\!\!\!\!6}{5} sen\bigg(\dfrac{5x}{6}\bigg)+\dfrac{1}{\backslash\!\!\!\!2}\cdot\backslash\!\!\!6 sen\bigg(\dfrac{x}{6}\bigg)+k\\\\\displaystyle\sf\int cos\bigg(\dfrac{x}{2}\bigg)~cos\bigg(\dfrac{x}{3}\bigg)dx=\dfrac{3}{5} sen\bigg(\dfrac{5x}{6}\bigg)+3 sen\bigg(\dfrac{x}{6}\bigg)+k\end{array}}$