int_0^{2 pi} theta sin ^6 theta cos theta , d | vedclass

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(D) Let $I = \int_0^{2 \pi} 	heta \sin ^6 	heta \cos 	heta \, d	heta \quad \dots (1)$ Applying the property $\int_0^a f(x) \, dx = \int_0^a f(a-x)

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(D) Let $I = \int_0^{2 \pi} 	heta \sin ^6 	heta \cos 	heta \, d	heta \quad \dots (1)$ Applying the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$, we get: $I = \int_0^{2 \pi} (2 \pi - 	heta) \sin ^6 (2 \pi - 	heta) \cos (2 \pi - 	heta) \, d	heta$ Since $\sin(2 \pi - 	heta) = -\sin 	heta$ and $\cos(2 \pi - 	heta) = \cos 	heta$, we have: $I = \int_0^{2 \pi} (2 \pi - 	heta) \sin ^6 	heta \cos 	heta \, d	heta$ $I = 2 \pi \int_0^{2 \pi} \sin ^6 	heta \cos 	heta \, d	heta - \int_0^{2 \pi} 	heta \sin ^6 	heta \cos 	heta \, d	heta$ $I = 2 \pi \int_0^{2 \pi} \sin ^6 	heta \cos 	heta \, d	heta - I$ $2I = 2 \pi \int_0^{2 \pi} \sin ^6 	heta \cos 	heta \, d	heta$ $I = \pi \int_0^{2 \pi} \sin ^6 	heta \cos 	heta \, d	heta$ Let $f(	heta) = \sin ^6 	heta \cos 	heta$. Then $f(2 \pi - 	heta) = \sin ^6 (2 \pi - 	heta) \cos (2 \pi - 	heta) = (-\sin 	heta)^6 \cos 	heta = \sin ^6 	heta \cos 	heta = f(	heta)$. Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$: $I = \pi \cdot 2 \int_0^{\pi} \sin ^6 	heta \cos 	heta \, d	heta = 2 \pi \int_0^{\pi} \sin ^6 	heta \cos 	heta \, d	heta$ Let $u = \sin 	heta$, then $du = \cos 	heta \, d	heta$. When $	heta = 0, u = 0$. When $	heta = \pi, u = 0$. $I = 2 \pi \int_0^0 u^6 \, du = 0$.

$\int_0^{2 \pi} \theta \sin ^6 \theta \cos \theta \, d\theta$ is equal to

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