int_0^a {x^4 sqrt{a^2 - x^2}} ,dx = | vedclass

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Question

(B) Let $I = \int_0^a x^4 \sqrt{a^2 - x^2} \,dx$. Put $x = a \sin 	heta$,then $dx = a \cos 	heta \,d	heta$. When $x = 0$,$	heta = 0$. When $x = a$

Vedclass · Accepted Answer

$\frac{\pi}{32} a^6$

Vedclass · Answer

(B) Let $I = \int_0^a x^4 \sqrt{a^2 - x^2} \,dx$. Put $x = a \sin 	heta$,then $dx = a \cos 	heta \,d	heta$. When $x = 0$,$	heta = 0$. When $x = a$,$	heta = \frac{\pi}{2}$. Substituting these into the integral: $I = \int_0^{\pi/2} (a \sin 	heta)^4 \sqrt{a^2 - a^2 \sin^2 	heta} \cdot (a \cos 	heta) \,d	heta$ $I = \int_0^{\pi/2} a^4 \sin^4 	heta \cdot (a \cos 	heta) \cdot (a \cos 	heta) \,d	heta$ $I = a^6 \int_0^{\pi/2} \sin^4 	heta \cos^2 	heta \,d	heta$. Using the reduction formula $\int_0^{\pi/2} \sin^m 	heta \cos^n 	heta \,d	heta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$: $I = a^6 \cdot \frac{\Gamma(\frac{5}{2}) \Gamma(\frac{3}{2})}{2 \Gamma(\frac{4+2+2}{2})} = a^6 \cdot \frac{(\frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi}) \cdot (\frac{1}{2} \cdot \sqrt{\pi})}{2 \cdot \Gamma(4)}$ $I = a^6 \cdot \frac{\frac{3}{8} \pi}{2 \cdot 6} = a^6 \cdot \frac{3\pi}{96} = \frac{\pi}{32} a^6$.

$\int_0^a {x^4 \sqrt{a^2 - x^2}} \,dx = $

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