$\int\left(\sin ^{3} 2 \theta+5\right) \cos 2 \theta d \theta=\frac{1}{4} \sin ^{4} 2 \theta+\frac{5}{2} \sin 2 \theta+C$ .

Name: $\int\left(\sin ^{3} 2 \theta+5\right) \cos 2 \theta d \theta=\frac{1}{4} \sin ^{4} 2 \theta+\frac{5}{2} \sin 2 \theta+C$ .
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Question

Quizplus · Accepted Answer

The integral of $\sin^n x \cos x dx$ typically results in a form involving $\sin^{n+1} x$, but the constant term $5$ when integrated with respect to $\theta$ should result in $5\theta$, not $\frac{5}{2} \sin 2 \theta$.

∫(sin⁡32θ+5)cos⁡2θdθ=14sin⁡42θ+52sin⁡2θ+C\int\left(\sin ^{3} 2 \theta+5\right) \cos 2 \theta d \theta=\frac{1}{4} \sin ^{4} 2 \theta+\frac{5}{2} \sin 2 \theta+C∫(sin32θ+5)cos2θdθ=41​sin42θ+25​sin2θ+C

$\int\left(\sin ^{3} 2 \theta+5\right) \cos 2 \theta d \theta=\frac{1}{4} \sin ^{4} 2 \theta+\frac{5}{2} \sin 2 \theta+C$