int limits_{0}^{pi} frac{e^{cos x} sin x}{lef | vedclass

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

Question

(C) माना $I = \int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{(1+\cos^2 x)(e^{\cos x}+e^{-\cos x})} dx$  $(1)$गुणधर्म $\int_{0}^{a} f(x) dx = \int_{0}

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) माना $I = \int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{(1+\cos^2 x)(e^{\cos x}+e^{-\cos x})} dx$  $(1)$गुणधर्म $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ का उपयोग करने पर:$I = \int \limits_{0}^{\pi} \frac{e^{\cos(\pi-x)} \sin(\pi-x)}{(1+\cos^2(\pi-x))(e^{\cos(\pi-x)}+e^{-\cos(\pi-x)})} dx$$I = \int \limits_{0}^{\pi} \frac{e^{-\cos x} \sin x}{(1+\cos^2 x)(e^{-\cos x}+e^{\cos x})} dx$  $(2)$$(1)$ और $(2)$ को जोड़ने पर:$2I = \int \limits_{0}^{\pi} \frac{(e^{\cos x} + e^{-\cos x}) \sin x}{(1+\cos^2 x)(e^{\cos x}+e^{-\cos x})} dx$$2I = \int \limits_{0}^{\pi} \frac{\sin x}{1+\cos^2 x} dx$चूंकि $\frac{\sin x}{1+\cos^2 x}$ फलन $x = \frac{\pi}{2}$ के सापेक्ष सममित है,इसलिए $2I = 2 \int \limits_{0}^{\pi/2} \frac{\sin x}{1+\cos^2 x} dx$$I = \int \limits_{0}^{\pi/2} \frac{\sin x}{1+\cos^2 x} dx$माना $\cos x = t$,तो $-\sin x dx = dt$। जब $x=0, t=1$; जब $x=\pi/2, t=0$।$I = -\int \limits_{1}^{0} \frac{dt}{1+t^2} = \int \limits_{0}^{1} \frac{dt}{1+t^2} = [	an^{-1} t]_{0}^{1} = 	an^{-1}(1) - 	an^{-1}(0) = \frac{\pi}{4}$

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

Similar Questions

$\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cdot \cos x} d x$ का मान ज्ञात कीजिए।

यदि $I_n = \int_0^{\pi / 2} \sin^n(x) dx$ और $I_n = (k) I_{n-2}$ है,तो $k$ का मान क्या होगा?

$\int_0^1 (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} dx$ का मान किसके बराबर है?

Vedclass Test Series

Exam Paper Generator

Online Exam Module