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(C) Let $I = \int_0^\pi x \sin x \cos^4 x \, dx \quad \dots(1)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$I = \int_0^\pi

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$\frac{\pi}{5}$

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(C) Let $I = \int_0^\pi x \sin x \cos^4 x \, dx \quad \dots(1)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:$I = \int_0^\pi (\pi - x) \sin(\pi - x) \cos^4(\pi - x) \, dx$$I = \int_0^\pi (\pi - x) \sin x (-\cos x)^4 \, dx$$I = \int_0^\pi (\pi - x) \sin x \cos^4 x \, dx \quad \dots(2)$Adding $(1)$ and $(2)$:$2I = \int_0^\pi (x + \pi - x) \sin x \cos^4 x \, dx$$2I = \pi \int_0^\pi \sin x \cos^4 x \, dx$Let $t = \cos x$,then $dt = -\sin x \, dx$. When $x=0, t=1$ and when $x=\pi, t=-1$.$2I = \pi \int_1^{-1} t^4 (-dt) = \pi \int_{-1}^1 t^4 \, dt$Since $t^4$ is an even function,$\int_{-1}^1 t^4 \, dt = 2 \int_0^1 t^4 \, dt$.$2I = 2\pi \left[ \frac{t^5}{5} \right]_0^1 = 2\pi \left( \frac{1}{5} \right) = \frac{2\pi}{5}$$I = \frac{\pi}{5}$

$\int_0^\pi x \sin x \cos^4 x \, dx = $

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