The value of int_{-3pi}^{3pi} sin^2 theta sin | vedclass

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Question

(B) Let $I = \int_{-3\pi}^{3\pi} \sin^2 	heta \sin^2 2	heta \, d	heta$. Since the integrand is an even function,$I = 2 \int_{0}^{3\pi} \sin^2 	het

Vedclass · Accepted Answer

$\frac{3\pi}{2}$

Vedclass · Answer

(B) Let $I = \int_{-3\pi}^{3\pi} \sin^2 	heta \sin^2 2	heta \, d	heta$. Since the integrand is an even function,$I = 2 \int_{0}^{3\pi} \sin^2 	heta \sin^2 2	heta \, d	heta$.Using $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have $\sin^2 2	heta = 4 \sin^2 	heta \cos^2 	heta$.So,$I = 2 \int_{0}^{3\pi} \sin^2 	heta (4 \sin^2 	heta \cos^2 	heta) \, d	heta = 8 \int_{0}^{3\pi} \sin^4 	heta \cos^2 	heta \, d	heta$.Since $\sin^4 	heta \cos^2 	heta$ has a period of $\pi$,$\int_{0}^{3\pi} f(	heta) \, d	heta = 3 \int_{0}^{\pi} f(	heta) \, d	heta$.Thus,$I = 8 	imes 3 \int_{0}^{\pi} \sin^4 	heta \cos^2 	heta \, d	heta = 24 \int_{0}^{\pi} \sin^4 	heta \cos^2 	heta \, d	heta$.Using the property $\int_{0}^{\pi} f(\sin 	heta) \, d	heta = 2 \int_{0}^{\pi/2} f(\sin 	heta) \, d	heta$,we get $I = 48 \int_{0}^{\pi/2} \sin^4 	heta \cos^2 	heta \, d	heta$.Using Wallis's formula $\int_{0}^{\pi/2} \sin^m 	heta \cos^n 	heta \, d	heta = \frac{(m-1)!!(n-1)!!}{(m+n)!!} 	imes \frac{\pi}{2}$ (for even $m, n$):$I = 48 	imes \frac{(3 	imes 1) 	imes (1)}{(6 	imes 4 	imes 2)} 	imes \frac{\pi}{2} = 48 	imes \frac{3}{48} 	imes \frac{\pi}{2} = \frac{3\pi}{2}$.

The value of $\int_{-3\pi}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$ is equal to:

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