The value of int_{0}^{pi} |sin^3 theta| , dth | vedclass

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(C) Let $I = \int_{0}^{\pi} |\sin^3 	heta| \, d	heta$. Since $\sin 	heta \ge 0$ for all $	heta \in [0, \pi]$,we have $|\sin^3 	heta| = \sin^3 	h

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$4/3$

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(C) Let $I = \int_{0}^{\pi} |\sin^3 	heta| \, d	heta$. Since $\sin 	heta \ge 0$ for all $	heta \in [0, \pi]$,we have $|\sin^3 	heta| = \sin^3 	heta$. Therefore,$I = \int_{0}^{\pi} \sin^3 	heta \, d	heta$. Using the identity $\sin^3 	heta = \sin 	heta (1 - \cos^2 	heta)$,we get: $I = \int_{0}^{\pi} \sin 	heta (1 - \cos^2 	heta) \, d	heta = \int_{0}^{\pi} \sin 	heta \, d	heta - \int_{0}^{\pi} \sin 	heta \cos^2 	heta \, d	heta$. Evaluating the integrals: $\int_{0}^{\pi} \sin 	heta \, d	heta = [-\cos 	heta]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$. For the second part,let $u = \cos 	heta$,then $du = -\sin 	heta \, d	heta$. When $	heta = 0, u = 1$; when $	heta = \pi, u = -1$. $-\int_{0}^{\pi} \sin 	heta \cos^2 	heta \, d	heta = \int_{1}^{-1} u^2 \, du = [\frac{u^3}{3}]_{1}^{-1} = \frac{(-1)^3}{3} - \frac{1^3}{3} = -\frac{1}{3} - \frac{1}{3} = -\frac{2}{3}$. Thus,$I = 2 - \frac{2}{3} = \frac{4}{3}$.

The value of $\int_{0}^{\pi} |\sin^3 \theta| \, d\theta$ is

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