Question

Encontrar a integral de:
∫cos(4x)*sen(2x) dx

Answer 1

Fórmulas importantes:

$\boxed{\boxed{sen^{2}(\theta)+cos^{2}(\theta)=1~~\forall~\theta\in\mathbb{R}}}\\\\\\\boxed{\boxed{sen(\alpha\pm\beta)=sen(\alpha)cos(\beta)\pm sen(\beta)cos(\alpha)}}$
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Note que

$sen(4x+2x)=sen(4x)cos(2x)+sen(2x)cos(4x)\\\\sen(4x-2x)=sen(4x)cos(2x)-sen(2x)cos(4x)$

Suntraindo as duas equações:

$sen(4x+2x)-sen(4x-2x)=sen(2x)cos(4x)+sen(2x)cos(4x)\\\\sen(6x)-sen(2x)=2\cdot cos(4x)sen(2x)$

Então:

$\boxed{\boxed{cos(4x)sen(2x)=\dfrac{1}{2}\bigg[sen(6x)-sen(2x)\bigg]}}$
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$\displaystyle\int cos(4x)sen(2x)\,dx=\int\dfrac{1}{2}\bigg[sen(6x)-sen(2x)\bigg]\,dx\\\\\\\int cos(4x)sen(2x)\,dx=\dfrac{1}{2}\bigg[\int sen(6x)\,dx-\int sen(2x)\,dx\bigg]$

Essas integrais são fáceis:

Fazendo $\begin{cases}u=6x~\rightarrow~du=6dx~\rightarrow~dx=\frac{1}{6}du\\\\a=2x~\rightarrow~da=2da~\rightarrow~dx=\frac{1}{2}da\end{cases}$

temos

$\displaystyle\int sen(6x)\,dx=\int sen(u)\dfrac{1}{6}\,du=\dfrac{1}{6}\int sen(u)\,du=-\dfrac{1}{6}\,cos(6x)\\\\\\\int sen(2x)\,dx=\int sen(a)\dfrac{1}{2}\,da=\dfrac{1}{2}\int sen(a)\,du=-\dfrac{1}{2}\,cos(2x)$

(não adicionei constantes)

Portanto:

$\displaystyle\int cos(4x)sen(2x)\,dx=\dfrac{1}{2}\bigg[-\dfrac{1}{6}\,cos(6x)+\dfrac{1}{2}\,cos(2x)\bigg]+C\\\\\\\boxed{\boxed{\int cos(4x)sen(2x)\,dx=\dfrac{1}{4}\,cos(2x)-\dfrac{1}{12}\,cos(6x)+C}}$