समाकलन int limits_{-frac{pi}{4}}^{frac{pi}{4} | vedclass

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

Question

(D) माना $I = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$  $(1)$गुणधर्म $\int_a^b f(x) dx = \int_a^b f(a+b-x

Vedclass · Accepted Answer

$\frac{\pi^2}{6 \sqrt{3}}$

Vedclass · Answer

(D) माना $I = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$  $(1)$गुणधर्म $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ का उपयोग करते हुए,जहाँ $a+b = 0$,हम $x$ को $-x$ से प्रतिस्थापित करते हैं:$I = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-x+\frac{\pi}{4}}{2-\cos 2(-x)} d x = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-x+\frac{\pi}{4}}{2-\cos 2 x} d x$  $(2)$$(1)$ और $(2)$ को जोड़ने पर:$2I = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{(x+\frac{\pi}{4}) + (-x+\frac{\pi}{4})}{2-\cos 2 x} d x = \int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{2}}{2-\cos 2 x} d x$चूँकि फलन सम है,$2I = 2 \cdot \frac{\pi}{2} \int \limits_0^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x = \pi \int \limits_0^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x$$I = \frac{\pi}{2} \int \limits_0^{\frac{\pi}{4}} \frac{\sec^2 x}{2(1+	an^2 x) - (1-	an^2 x)} dx = \frac{\pi}{2} \int \limits_0^{\frac{\pi}{4}} \frac{\sec^2 x}{1+3	an^2 x} dx$माना $t = 	an x$,तब $dt = \sec^2 x dx$. जब $x=0, t=0$; जब $x=\frac{\pi}{4}, t=1$:$I = \frac{\pi}{2} \int \limits_0^1 \frac{dt}{1+3t^2} = \frac{\pi}{2 \cdot 3} \int \limits_0^1 \frac{dt}{\frac{1}{3}+t^2} = \frac{\pi}{6} \cdot \sqrt{3} [	an^{-1}(\sqrt{3}t)]_0^1 = \frac{\pi \sqrt{3}}{6} 	an^{-1}(\sqrt{3}) = \frac{\pi \sqrt{3}}{6} \cdot \frac{\pi}{3} = \frac{\pi^2}{6 \sqrt{3}}$

समाकलन $\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$ का मान ज्ञात कीजिए :

Similar Questions

यदि $a > 0$ है, तो $\int_{-\pi}^{\pi} \frac{\sin^2 x}{1+a^x} dx$ का मान ज्ञात कीजिए।

$\int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{\left(1+\cos ^{2} x\right)\left(e^{\cos x}+e^{-\cos x}\right)} d x$ का मान ज्ञात कीजिए।

$\int_0^{2 \pi} \sin ^3 x \cos ^2 x \, dx = $ . . . . . . .

$\int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx = $

$\int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module