int_0^pi x cdot sin^5 x cdot cos^6 x , dx = | vedclass

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(B) Let $I = \int_0^\pi x \sin^5 x \cos^6 x \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^\pi (\pi - x)

Vedclass · Accepted Answer

$\frac{8 \pi}{693}$

Vedclass · Answer

(B) Let $I = \int_0^\pi x \sin^5 x \cos^6 x \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^\pi (\pi - x) \sin^5(\pi - x) \cos^6(\pi - x) \, dx$ Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we have: $I = \int_0^\pi (\pi - x) \sin^5 x (-\cos x)^6 \, dx = \int_0^\pi (\pi - x) \sin^5 x \cos^6 x \, dx$ $I = \pi \int_0^\pi \sin^5 x \cos^6 x \, dx - I$ $2I = \pi \int_0^\pi \sin^5 x \cos^6 x \, dx$ Since $\sin^5 x \cos^6 x$ is symmetric about $x = \pi/2$,$\int_0^\pi \sin^5 x \cos^6 x \, dx = 2 \int_0^{\pi/2} \sin^5 x \cos^6 x \, dx$. Using Wallis' formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!}$: $\int_0^{\pi/2} \sin^5 x \cos^6 x \, dx = \frac{(4 \cdot 2) \cdot (5 \cdot 3 \cdot 1)}{11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1} = \frac{8 \cdot 15}{11 \cdot 9 \cdot 7 \cdot 15} = \frac{8}{693}$. Thus,$2I = \pi \cdot 2 \cdot \frac{8}{693} = \frac{16\pi}{693}$. $I = \frac{8\pi}{693}$.

$\int_0^\pi x \cdot \sin^5 x \cdot \cos^6 x \, dx =$

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