If I_n = int_0^a frac{x^n}{sqrt{a^2-x^2}} dx, | vedclass

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(B) We have, $I_n = \int_0^a \frac{x^n}{\sqrt{a^2-x^2}} dx$. Substitute $x = a \sin 	heta$, then $dx = a \cos 	heta d	heta$. When $x = 0$, $	heta

Vedclass · Accepted Answer

$\frac{35}{48} a^4$

Vedclass · Answer

(B) We have, $I_n = \int_0^a \frac{x^n}{\sqrt{a^2-x^2}} dx$. Substitute $x = a \sin 	heta$, then $dx = a \cos 	heta d	heta$. When $x = 0$, $	heta = 0$ and when $x = a$, $	heta = \frac{\pi}{2}$. Thus, $I_n = \int_0^{\pi/2} \frac{a^n \sin^n 	heta}{a \cos 	heta} \cdot a \cos 	heta d	heta = a^n \int_0^{\pi/2} \sin^n 	heta d	heta$. Using Wallis' formula, $\int_0^{\pi/2} \sin^n 	heta d	heta = \frac{(n-1)(n-3)\dots(1)}{n(n-2)\dots(2)} \cdot \frac{\pi}{2}$ for even $n$. $I_8 = a^8 \int_0^{\pi/2} \sin^8 	heta d	heta = a^8 \left( \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} ight)$. $I_4 = a^4 \int_0^{\pi/2} \sin^4 	heta d	heta = a^4 \left( \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} ight)$. Therefore, $\frac{I_8}{I_4} = \frac{a^8 \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}}{a^4 \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}} = a^4 \cdot \frac{7}{8} \cdot \frac{5}{6} = \frac{35}{48} a^4$.

If $I_n = \int_0^a \frac{x^n}{\sqrt{a^2-x^2}} dx$, then $\frac{I_8}{I_4} =$

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