int_{-3pi}^{3pi} sin^2 theta sin^2 2theta , d | vedclass

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

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 $f(	heta) = \sin^2 	heta \sin^2 2	heta$ 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 the period of $\sin^4 	heta \cos^2 	heta$ is $\pi$,$\int_{0}^{3\pi} f(	heta) \, d	heta = 3 \int_{0}^{\pi} f(	heta) \, d	heta$. Thus,$I = 8 	imes 3 \int_{0}^{\pi} \sin^4 	heta \cos^2 	heta \, d	heta = 24 	imes 2 \int_{0}^{\pi/2} \sin^4 	heta \cos^2 	heta \, d	heta = 48 \int_{0}^{\pi/2} \sin^4 	heta \cos^2 	heta \, d	heta$. 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 $m, n$): $I = 48 	imes \frac{(3 	imes 1) 	imes (1)}{(6 	imes 4 	imes 2)} 	imes \frac{\pi}{2} = 48 	imes \frac{3}{48} 	imes \frac{\pi}{2} = \frac{3\pi}{2}$.

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

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