int_{-pi/2}^{pi/2} sin^2 x cos^2 x (sin x + c | vedclass

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Question

(B) Let $I = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx$. Split the integral into two parts: $I = \int_{-\pi/2}^{\pi/2} \sin^3 x

Vedclass · Accepted Answer

$\frac{4}{15}$

Vedclass · Answer

(B) Let $I = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx$. Split the integral into two parts: $I = \int_{-\pi/2}^{\pi/2} \sin^3 x \cos^2 x \, dx + \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^3 x \, dx$. The first integral $\int_{-\pi/2}^{\pi/2} \sin^3 x \cos^2 x \, dx$ is an odd function,so its value is $0$. The second integral $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^3 x \, dx$ is an even function,so it equals $2 \int_0^{\pi/2} \sin^2 x \cos^3 x \, dx$. Using Wallis' formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!}$ for even $m, n$: $2 \int_0^{\pi/2} \sin^2 x \cos^3 x \, dx = 2 	imes \frac{(2-1)!!(3-1)!!}{(2+3)!!} = 2 	imes \frac{1 	imes 2}{5 	imes 3 	imes 1} = 2 	imes \frac{2}{15} = \frac{4}{15}$.

$\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx = $

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