Let [bullet] be the greatest integer function | vedclass

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(B) First, we calculate $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx = \int_{0}^{64} x^{1/3} dx - \int_{0}^{64} [x^{1/3}] dx$. $\int_{0}^{64} x^{1

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) First, we calculate $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx = \int_{0}^{64} x^{1/3} dx - \int_{0}^{64} [x^{1/3}] dx$. $\int_{0}^{64} x^{1/3} dx = \left[ \frac{3}{4} x^{4/3} ight]_{0}^{64} = \frac{3}{4} 	imes 256 = 192$. For $\int_{0}^{64} [x^{1/3}] dx$, we split the integral based on the values of $[x^{1/3}]$: $= \int_{0}^{1} 0 dx + \int_{1}^{8} 1 dx + \int_{8}^{27} 2 dx + \int_{27}^{64} 3 dx = 0 + (8-1) + 2(27-8) + 3(64-27) = 7 + 38 + 111 = 156$. Thus, $\alpha = 192 - 156 = 36$. Now, we evaluate $E = \frac{1}{\pi} \int_{0}^{36\pi} \frac{\sin^2 	heta}{\sin^6 	heta + \cos^6 	heta} d	heta$. Since the integrand has a period of $\pi$, $E = \frac{36}{\pi} \int_{0}^{\pi} \frac{\sin^2 	heta}{\sin^6 	heta + \cos^6 	heta} d	heta = \frac{36 	imes 2}{\pi} \int_{0}^{\pi/2} \frac{\sin^2 	heta}{\sin^6 	heta + \cos^6 	heta} d	heta$. Let $J = \int_{0}^{\pi/2} \frac{\sin^2 	heta}{\sin^6 	heta + \cos^6 	heta} d	heta$. By King's property, $J = \int_{0}^{\pi/2} \frac{\cos^2 	heta}{\sin^6 	heta + \cos^6 	heta} d	heta$. Adding these, $2J = \int_{0}^{\pi/2} \frac{1}{\sin^6 	heta + \cos^6 	heta} d	heta = \int_{0}^{\pi/2} \frac{\sec^6 	heta}{	an^6 	heta + 1} d	heta$. Let $	an 	heta = t$, then $dt = \sec^2 	heta d	heta$. $2J = \int_{0}^{\infty} \frac{(1+t^2)^2}{t^6+1} dt = \int_{0}^{\infty} \frac{1+t^2}{t^4-t^2+1} dt = \pi$. Thus $J = \pi/2$. Finally, $E = \frac{72}{\pi} 	imes \frac{\pi}{2} = 36$.

Let $[\bullet]$ be the greatest integer function. If $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx$, then $\frac{1}{\pi} \int_{0}^{\alpha\pi} \left( \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} \right) d\theta$ is equal to . . . . . . .

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