int_0^{2 pi} sin^6 x cos^5 x , dx is equal to | vedclass

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(C) Let $I = \int_0^{2 \pi} \sin^6 x \cos^5 x \, dx$. Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$: Here,$f

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(C) Let $I = \int_0^{2 \pi} \sin^6 x \cos^5 x \, dx$. Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$: Here,$f(x) = \sin^6 x \cos^5 x$. $f(2 \pi - x) = \sin^6(2 \pi - x) \cos^5(2 \pi - x) = (-\sin x)^6 (\cos x)^5 = \sin^6 x \cos^5 x = f(x)$. So,$I = 2 \int_0^{\pi} \sin^6 x \cos^5 x \, dx$. Now,using the property $\int_0^a f(x) \, dx = 0$ if $f(a-x) = -f(x)$: Let $g(x) = \sin^6 x \cos^5 x$. $g(\pi - x) = \sin^6(\pi - x) \cos^5(\pi - x) = (\sin x)^6 (-\cos x)^5 = -\sin^6 x \cos^5 x = -g(x)$. Since $g(\pi - x) = -g(x)$,the integral $\int_0^{\pi} \sin^6 x \cos^5 x \, dx = 0$. Therefore,$I = 2 	imes 0 = 0$.

$\int_0^{2 \pi} \sin^6 x \cos^5 x \, dx$ is equal to

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