The value of int_0^{2pi} |sin^3 theta| , dthe | vedclass

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(C) The function $f(	heta) = |\sin^3 	heta|$ is periodic with period $\pi$. Thus,$\int_0^{2\pi} |\sin^3 	heta| \, d	heta = 2 \int_0^{\pi} |\sin^3

Vedclass · Accepted Answer

$8/3$

Vedclass · Answer

(C) The function $f(	heta) = |\sin^3 	heta|$ is periodic with period $\pi$. Thus,$\int_0^{2\pi} |\sin^3 	heta| \, d	heta = 2 \int_0^{\pi} |\sin^3 	heta| \, d	heta$. Since $\sin 	heta \ge 0$ for $	heta \in [0, \pi]$,we have $|\sin^3 	heta| = \sin^3 	heta$. Using the identity $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$,we get $\sin^3 	heta = \frac{3\sin 	heta - \sin 3	heta}{4}$. Therefore,the integral becomes $2 \int_0^{\pi} \frac{3\sin 	heta - \sin 3	heta}{4} \, d	heta = \frac{1}{2} \int_0^{\pi} (3\sin 	heta - \sin 3	heta) \, d	heta$. Evaluating the integral: $\frac{1}{2} [-3\cos 	heta + \frac{1}{3}\cos 3	heta]_0^{\pi}$. $= \frac{1}{2} [(-3\cos \pi + \frac{1}{3}\cos 3\pi) - (-3\cos 0 + \frac{1}{3}\cos 0)]$. $= \frac{1}{2} [(3 - 1/3) - (-3 + 1/3)] = \frac{1}{2} [8/3 - (-8/3)] = \frac{1}{2} [16/3] = 8/3$.

The value of $\int_0^{2\pi} |\sin^3 \theta| \, d\theta$ is

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