(A) Let $I = \int_0^\pi \sin^2 x \cos^3 x \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have: $I = \int_0^\pi \sin^2(\pi

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(A) Let $I = \int_0^\pi \sin^2 x \cos^3 x \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have: $I = \int_0^\pi \sin^2(\pi - x) \cos^3(\pi - x) \, dx$. Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we get: $I = \int_0^\pi \sin^2 x (-\cos x)^3 \, dx = -\int_0^\pi \sin^2 x \cos^3 x \, dx$. Thus,$I = -I$,which implies $2I = 0$,so $I = 0$.

Class 12 Mathematics 7-2.Definite Integral Properties of definite integration

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$\int_0^\pi \sin^2 x \cos^3 x \, dx = $ . . . . . . .

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A
$0$
B
$-1$
C
$1$
D
$\pi$

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