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

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(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x$.We can split the integral into two parts: $I = \int_{-\pi / 2}^{\pi / 2

Vedclass · Accepted Answer

$\frac{4}{15}$

Vedclass · Answer

(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x$.We can split the integral into two parts: $I = \int_{-\pi / 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x + \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^3 x d x$.Let $f(x) = \sin ^3 x \cos ^2 x$. Since $f(-x) = \sin ^3(-x) \cos ^2(-x) = -\sin ^3 x \cos ^2 x = -f(x)$,$f(x)$ is an odd function. Thus,$\int_{-\pi / 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x = 0$.Now consider $g(x) = \sin ^2 x \cos ^3 x$. Since $g(-x) = \sin ^2(-x) \cos ^3(-x) = \sin ^2 x \cos ^3 x = g(x)$,$g(x)$ is an even function. Thus,$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^3 x d x = 2 \int_{0}^{\pi / 2} \sin ^2 x \cos ^3 x d x$.Using the reduction formula $\int_{0}^{\pi / 2} \sin^m x \cos^n x dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!}$ for even $m+n$,we have $2 \int_{0}^{\pi / 2} \sin^2 x \cos^3 x dx = 2 	imes \frac{(2-1)!!(3-1)!!}{(2+3+1)!!} = 2 	imes \frac{1 	imes 2}{6 	imes 4 	imes 2} = 2 	imes \frac{2}{48} = \frac{4}{48} = \frac{1}{12}$. Wait,let's re-evaluate: $2 \int_{0}^{\pi / 2} \sin^2 x \cos^2 x \cos x dx$. Let $u = \sin x$,then $du = \cos x dx$. When $x=0, u=0$; when $x=\pi/2, u=1$.$I = 2 \int_{0}^{1} u^2 (1-u^2) du = 2 \int_{0}^{1} (u^2 - u^4) du = 2 [\frac{u^3}{3} - \frac{u^5}{5}]_{0}^{1} = 2 (\frac{1}{3} - \frac{1}{5}) = 2 (\frac{2}{15}) = \frac{4}{15}$.

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

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