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$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x=$
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The correct answer is:
$\frac{4}{15}$
$\begin{aligned} & \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x \\ & =\int_{-\pi / 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x+\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^3 x d x \\ & =0+2 \int_0^{\pi / 2} \sin ^2 x \cos ^3 x d x=0+2 \times \frac{2}{15}=\frac{4}{15}\end{aligned}$
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