int_{-pi/6}^{pi/6} left( frac{pi + 4x^{11}}{1 | vedclass

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Question

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) माना $I = \int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx + \int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1 - \sin(|x| + \pi/6)} dx$.चूंकि $f(x) = \frac{4x^{11}}{1 - \sin(|x| + \pi/6)}$ एक विषम फलन है, इसलिए दूसरा समाकलन $0$ होगा।अतः, $I = 2 \int_{0}^{\pi/6} \frac{\pi}{1 - \sin(x + \pi/6)} dx = 2\pi \int_{0}^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$.माना $x + \pi/6 = t$, तो $dx = dt$। जब $x=0, t=\pi/6$; जब $x=\pi/6, t=\pi/3$।$I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{1 - \sin t} dt = 2\pi \int_{\pi/6}^{\pi/3} \frac{1 + \sin t}{\cos^2 t} dt$.$I = 2\pi \int_{\pi/6}^{\pi/3} (\sec^2 t + \sec t 	an t) dt = 2\pi [	an t + \sec t]_{\pi/6}^{\pi/3}$.$I = 2\pi [(	an(\pi/3) + \sec(\pi/3)) - (	an(\pi/6) + \sec(\pi/6))]$.$I = 2\pi [(\sqrt{3} + 2) - (1/\sqrt{3} + 2/\sqrt{3})] = 2\pi [\sqrt{3} + 2 - 3/\sqrt{3}] = 2\pi [\sqrt{3} + 2 - \sqrt{3}] = 4\pi$.अतः, मान $4$ है।

$\int_{-\pi/6}^{\pi/6} \left( \frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)} \right) dx$ का मान ज्ञात कीजिए: ($\pi$ में)

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