Answer
a) 2
b)2
Work Step by Step
a) let u=cos x =>du=-sin x dx=>-du=sin x dx;
x=0=>u=1,
x=$\pi$=>u=-1
$\int ^{\pi}_0 3\cos ^2x \sin xdx$
=$\int^{-1}_1 -3u^2 du$
=$[-u^3]^{-1}_1$
=$-(-1)^3-(-1(1)^3)=2$
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b)Use the same substitution as in part (a); x=$2\pi$=>u=1,
x=3$\pi$=>u=-1
$\int_{2\pi}^{3\pi}3cos^2x \sin x dx$
=$\int^{-1}_1 -3u^2du$
=2
