Question

por favor me ajudem a resolver está integral​

Answer 1

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$\Large\boxed{\underline{\sf sen(s)\cdot cos(t)=\dfrac{1}{2}[sen(s+t)+sen(s-t)]}}$

$\displaystyle\sf\int sen(2x)\cdot cos(4x)~dx=\dfrac{1}{2}\int[sen(2x+4x)+sen(2x-4x)]~dx\\\displaystyle\sf\dfrac{1}{2}\int sen(6x)~dx+\dfrac{1}{2}\int sen(-2x)~dx$

$\boxed{\displaystyle\sf\int sen(a~u)~du=-\dfrac{1}{a} cos(a~u)+k}$

$\displaystyle\sf\dfrac{1}{2}\int sen(6x)~dx=\dfrac{1}{2}\cdot\left(-\dfrac{1}{6}\right)cos (6x)+k=-\dfrac{1}{12}cos (6x)+k$

$\displaystyle\sf\dfrac{1}{2}\int sen (-2x)~dx=\dfrac{1}{2}\cdot\left(-\dfrac{1}{2}\right) cos(-2x)+k=-\dfrac{1}{4} cos(-2x)+k\\\sf =\dfrac{1}{4} cos(2x)+k\checkmark$

$\boxed{\displaystyle\sf\int sen(2x)\cdot cos(4x)~dx=-\dfrac{1}{12}cos(6x)+\dfrac{1}{4}cos(2x)+k}$