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Answer
$$ \displaystyle\int \dfrac{7{\cdot}\cos\left(6x\right)}{4{\cdot}{\left(\sin\left(6x\right)+4\right)}^{3}}\, \mathrm d x = -{{7}\over{48\,\left(\sin \left(6\,x\right)+4\right)^2}} $$
Explanation
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