int_0^{frac{pi}{2}} sin^4 theta cos^3 theta , | vedclass

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(B) Let $I = \int_0^{\frac{\pi}{2}} \sin^4 	heta \cos^3 	heta \, d	heta$. We can write $\cos^3 	heta \, d	heta$ as $\cos^2 	heta \cdot \cos 	he

Vedclass · Accepted Answer

$\frac{2}{35}$

Vedclass · Answer

(B) Let $I = \int_0^{\frac{\pi}{2}} \sin^4 	heta \cos^3 	heta \, d	heta$. We can write $\cos^3 	heta \, d	heta$ as $\cos^2 	heta \cdot \cos 	heta \, d	heta = (1 - \sin^2 	heta) \cos 	heta \, d	heta$. Substituting this into the integral, we get: $I = \int_0^{\frac{\pi}{2}} \sin^4 	heta (1 - \sin^2 	heta) \cos 	heta \, d	heta$. Let $t = \sin 	heta$, then $dt = \cos 	heta \, d	heta$. When $	heta = 0$, $t = 0$. When $	heta = \frac{\pi}{2}$, $t = 1$. $I = \int_0^1 t^4 (1 - t^2) \, dt = \int_0^1 (t^4 - t^6) \, dt$. Integrating term by term: $I = \left[ \frac{t^5}{5} - \frac{t^7}{7} ight]_0^1 = \left( \frac{1}{5} - \frac{1}{7} ight) - (0 - 0) = \frac{7 - 5}{35} = \frac{2}{35}$.

$\int_0^{\frac{\pi}{2}} \sin^4 \theta \cos^3 \theta \, d\theta =$

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