int_0^{pi /4} sec^7 theta sin^3 theta , dthet | vedclass

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(C) We are given the integral $I = \int_0^{\pi /4} \sec^7 	heta \sin^3 	heta \, d	heta$. Rewrite the integrand as: $I = \int_0^{\pi /4} \frac{\sin^

Vedclass · Accepted Answer

$\frac{5}{12}$

Vedclass · Answer

(C) We are given the integral $I = \int_0^{\pi /4} \sec^7 	heta \sin^3 	heta \, d	heta$. Rewrite the integrand as: $I = \int_0^{\pi /4} \frac{\sin^3 	heta}{\cos^3 	heta} \cdot \sec^4 	heta \, d	heta = \int_0^{\pi /4} 	an^3 	heta \sec^4 	heta \, d	heta$. Since $\sec^4 	heta = \sec^2 	heta \cdot \sec^2 	heta = (1 + 	an^2 	heta) \sec^2 	heta$,the integral becomes: $I = \int_0^{\pi /4} 	an^3 	heta (1 + 	an^2 	heta) \sec^2 	heta \, d	heta$. Let $t = 	an 	heta$,then $dt = \sec^2 	heta \, d	heta$. When $	heta = 0, t = 0$. When $	heta = \pi /4, t = 1$. Substituting these into the integral: $I = \int_0^1 t^3 (1 + t^2) \, dt = \int_0^1 (t^3 + t^5) \, dt$. Integrating term by term: $I = \left[ \frac{t^4}{4} + \frac{t^6}{6} ight]_0^1 = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}$.

$\int_0^{\pi /4} \sec^7 \theta \sin^3 \theta \, d\theta = $

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