int_0^pi x sin^4 x cos^6 x , dx = | vedclass

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(A) Let $I = \int_0^\pi x \sin^4 x \cos^6 x \, dx$. 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

$\frac{3 \pi^2}{512}$

Vedclass · Answer

(A) Let $I = \int_0^\pi x \sin^4 x \cos^6 x \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^\pi (\pi - x) \sin^4(\pi - x) \cos^6(\pi - x) \, dx$ Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we have: $I = \int_0^\pi (\pi - x) \sin^4 x (-\cos x)^6 \, dx = \int_0^\pi (\pi - x) \sin^4 x \cos^6 x \, dx$ $I = \pi \int_0^\pi \sin^4 x \cos^6 x \, dx - I$ $2I = \pi \int_0^\pi \sin^4 x \cos^6 x \, dx$ Using $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$: $2I = 2\pi \int_0^{\pi/2} \sin^4 x \cos^6 x \, dx$ $I = \pi \int_0^{\pi/2} \sin^4 x \cos^6 x \, dx$ Using Wallis' formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \cdot \frac{\pi}{2}$ (for even $m, n$): $I = \pi \left( \frac{3 \cdot 1 \cdot 5 \cdot 3 \cdot 1}{10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} \right) = \pi \left( \frac{45}{3840} \cdot \frac{\pi}{2} \right) = \pi \left( \frac{3}{256} \cdot \frac{\pi}{2} \right) = \frac{3 \pi^2}{512}$.

$\int_0^\pi x \sin^4 x \cos^6 x \, dx =$

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