int_0^2 x^8left(frac{4}{x^2}-1right)^{5 / 2} | vedclass

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(D) Let $I = \int_0^2 x^8 \left(\frac{4}{x^2} - 1ight)^{5/2} dx$. Substitute $x = 2 \sin 	heta$,so $dx = 2 \cos 	heta d	heta$. When $x = 0$,$	he

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$\frac{2^{10}}{63}$

Vedclass · Answer

(D) Let $I = \int_0^2 x^8 \left(\frac{4}{x^2} - 1ight)^{5/2} dx$. Substitute $x = 2 \sin 	heta$,so $dx = 2 \cos 	heta d	heta$. When $x = 0$,$	heta = 0$. When $x = 2$,$	heta = \frac{\pi}{2}$. The expression becomes $\frac{4}{x^2} - 1 = \frac{4}{4 \sin^2 	heta} - 1 = \csc^2 	heta - 1 = \cot^2 	heta$. Substituting these into the integral: $I = \int_0^{\pi/2} (2 \sin 	heta)^8 (\cot^2 	heta)^{5/2} (2 \cos 	heta) d	heta$ $I = \int_0^{\pi/2} (2^8 \sin^8 	heta) (\cot^5 	heta) (2 \cos 	heta) d	heta$ $I = 2^9 \int_0^{\pi/2} \sin^8 	heta \frac{\cos^5 	heta}{\sin^5 	heta} \cos 	heta d	heta$ $I = 2^9 \int_0^{\pi/2} \sin^3 	heta \cos^6 	heta d	heta$ Using the Wallis formula $\int_0^{\pi/2} \sin^m 	heta \cos^n 	heta d	heta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$: $I = 2^9 	imes \frac{\Gamma(2) \Gamma(7/2)}{2 \Gamma(11/2)} = 2^9 	imes \frac{1! 	imes \frac{5}{2} 	imes \frac{3}{2} 	imes \frac{1}{2} 	imes \sqrt{\pi}}{2 	imes \frac{9}{2} 	imes \frac{7}{2} 	imes \frac{5}{2} 	imes \frac{3}{2} 	imes \frac{1}{2} 	imes \sqrt{\pi}}$ $I = 2^9 	imes \frac{1}{2 	imes \frac{9}{2} 	imes \frac{7}{2}} = 2^9 	imes \frac{4}{63} = \frac{2^{11}}{63} = \frac{2048}{63}$. Wait,re-evaluating the integral: $I = 2^9 	imes \frac{2 	imes 1}{9 	imes 7 	imes 5} = \frac{512 	imes 2}{315} = \frac{1024}{315}$. Actually,calculating $\int_0^{\pi/2} \sin^3 	heta \cos^6 	heta d	heta = \int_0^{\pi/2} (1-\cos^2 	heta) \cos^6 	heta \sin 	heta d	heta$. Let $u = \cos 	heta$,$du = -\sin 	heta d	heta$. $I = 2^9 \int_0^1 (1-u^2) u^6 du = 2^9 [\frac{u^7}{7} - \frac{u^9}{9}]_0^1 = 512 (\frac{1}{7} - \frac{1}{9}) = 512 (\frac{2}{63}) = \frac{1024}{63} = \frac{2^{10}}{63}$.

$\int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x=$

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