The value of int_{-pi/6}^{pi/6} left( frac{pi | vedclass

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Question

(B) Let $I = \int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx + \int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1 - \sin(|x| + \pi/6)} dx$.Since $f(x)

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $I = \int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx + \int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1 - \sin(|x| + \pi/6)} dx$.Since $f(x) = \frac{4x^{11}}{1 - \sin(|x| + \pi/6)}$ is an odd function, the second integral is $0$.Thus, $I = 2 \int_{0}^{\pi/6} \frac{\pi}{1 - \sin(x + \pi/6)} dx = 2\pi \int_{0}^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$.Let $x + \pi/6 = t$, then $dx = dt$. When $x=0, t=\pi/6$; when $x=\pi/6, t=\pi/3$.$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{1 - \sin t} dt = 2\pi \int_{\pi/6}^{\pi/3} \frac{1 + \sin t}{\cos^2 t} dt$.$I = 2\pi \int_{\pi/6}^{\pi/3} (\sec^2 t + \sec t 	an t) dt = 2\pi [	an t + \sec t]_{\pi/6}^{\pi/3}$.$I = 2\pi [(	an(\pi/3) + \sec(\pi/3)) - (	an(\pi/6) + \sec(\pi/6))]$.$I = 2\pi [(\sqrt{3} + 2) - (1/\sqrt{3} + 2/\sqrt{3})] = 2\pi [\sqrt{3} + 2 - 3/\sqrt{3}] = 2\pi [\sqrt{3} + 2 - \sqrt{3}] = 4\pi$.Therefore, the value is $4$.

The value of $\int_{-\pi/6}^{\pi/6} \left( \frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)} \right) dx$ is equal to: (in $\pi$)

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