int_{-5 pi}^{5 pi} (1-cos 2x)^{frac{5}{2}} dx | vedclass

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(D) Let $I = \int_{-5 \pi}^{5 \pi} (1-\cos 2x)^{\frac{5}{2}} dx$. Since $f(x) = (1-\cos 2x)^{\frac{5}{2}}$ is an even function,$I = 2 \int_{0}^{5 \pi}

Vedclass · Accepted Answer

$\frac{128 \sqrt{2}}{3}$

Vedclass · Answer

(D) Let $I = \int_{-5 \pi}^{5 \pi} (1-\cos 2x)^{\frac{5}{2}} dx$. Since $f(x) = (1-\cos 2x)^{\frac{5}{2}}$ is an even function,$I = 2 \int_{0}^{5 \pi} (2 \sin^2 x)^{\frac{5}{2}} dx$. $I = 2 \int_{0}^{5 \pi} 2^{\frac{5}{2}} |\sin x|^5 dx = 2 	imes 4 \sqrt{2} \int_{0}^{5 \pi} |\sin x|^5 dx = 8 \sqrt{2} \int_{0}^{5 \pi} |\sin x|^5 dx$. Since $|\sin x|^5$ is periodic with period $\pi$,$\int_{0}^{5 \pi} |\sin x|^5 dx = 5 \int_{0}^{\pi} \sin^5 x dx = 5 	imes 2 \int_{0}^{\frac{\pi}{2}} \sin^5 x dx$. Using Wallis' formula,$\int_{0}^{\frac{\pi}{2}} \sin^5 x dx = \frac{4 	imes 2}{5 	imes 3} = \frac{8}{15}$. Thus,$I = 8 \sqrt{2} 	imes 5 	imes 2 	imes \frac{8}{15} = 80 \sqrt{2} 	imes \frac{8}{15} = 16 \sqrt{2} 	imes \frac{8}{3} = \frac{128 \sqrt{2}}{3}$.

$\int_{-5 \pi}^{5 \pi} (1-\cos 2x)^{\frac{5}{2}} dx =$

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