$\int \frac{1+\sin (4 x)}{4} d x=\frac{1}{4} x-\frac{1}{4} \cos (4 x)+C$ .

Name: $\int \frac{1+\sin (4 x)}{4} d x=\frac{1}{4} x-\frac{1}{4} \cos (4 x)+C$ .
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Question

Quizplus · Accepted Answer

The correct integral of $\frac{1+\sin(4x)}{4}$ is $\frac{1}{4}x - \frac{1}{16}\cos(4x) + C$, because when integrating $\sin(4x)$, the coefficient $4$ in the argument of sine function leads to a division by $4$ in the result of the integral of $\sin(4x)$, making the coefficient of $\cos(4x)$ in the result $\frac{1}{16}$ instead of $\frac{1}{4}$.

∫1+sin⁡(4x)4dx=14x−14cos⁡(4x)+C\int \frac{1+\sin (4 x)}{4} d x=\frac{1}{4} x-\frac{1}{4} \cos (4 x)+C∫41+sin(4x)​dx=41​x−41​cos(4x)+C

$\int \frac{1+\sin (4 x)}{4} d x=\frac{1}{4} x-\frac{1}{4} \cos (4 x)+C$