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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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)

Similar Questions

$\int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}\right] d x=$

$\int_{-7}^{7} \frac{5^x}{5^{[x]}} dx$ का मान ज्ञात कीजिए (जहाँ $[.]$ महत्तम पूर्णांक फलन को दर्शाता है)।

यदि $f(x) = \frac{x^3+5}{\sqrt{12+x}}$ और $\int_{-5}^5 f(x) dx = \int_0^5 (f(x) + g(x)) dx$ है,तो $g(x) =$

$\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+ e^{\sin x}} dx$ का मान ज्ञात कीजिए।

समाकलन $\int \limits_1^3 \left((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1\right) dx$ का मान है

Vedclass Test Series

Exam Paper Generator

Online Exam Module