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(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x) dx$ ...$(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+

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$\frac{4}{15}$

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(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x) dx$ ...$(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, where $a = -\frac{\pi}{2}$ and $b = \frac{\pi}{2}$, we have $a+b = 0$.$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2(-x) \cos^2(-x)(\sin(-x) + \cos(-x)) dx$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(-\sin x + \cos x) dx$ ...(ii)Adding $(i)$ and (ii):$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x - \sin x + \cos x) dx$$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(2 \cos x) dx$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^3 x dx$Since $\sin^2 x \cos^3 x$ is an even function, $I = 2 \int_0^{\frac{\pi}{2}} \sin^2 x \cos^3 x dx$.Let $\sin x = t$, then $\cos x dx = dt$. When $x=0, t=0$; when $x=\frac{\pi}{2}, t=1$.$I = 2 \int_0^1 t^2 (1-t^2) dt = 2 \int_0^1 (t^2 - t^4) dt$$I = 2 \left[ \frac{t^3}{3} - \frac{t^5}{5} \right]_0^1 = 2 \left( \frac{1}{3} - \frac{1}{5} \right) = 2 \left( \frac{2}{15} \right) = \frac{4}{15}$.

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

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