$\int \cos (6 \theta) d \theta=-\frac{1}{6} \sin (6 \theta)+C$ .

Name: $\int \cos (6 \theta) d \theta=-\frac{1}{6} \sin (6 \theta)+C$ .
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Question

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The correct integral of $\cos(6\theta)$ with respect to $\theta$ is $\frac{1}{6} \sin(6\theta) + C$, because the integral of $\cos(ax)$ is $\frac{1}{a}\sin(ax) + C$, where $a$ is a constant.

∫cos⁡(6θ)dθ=−16sin⁡(6θ)+C\int \cos (6 \theta) d \theta=-\frac{1}{6} \sin (6 \theta)+C∫cos(6θ)dθ=−61​sin(6θ)+C

$\int \cos (6 \theta) d \theta=-\frac{1}{6} \sin (6 \theta)+C$