The value of int limits_{0}^{pi} frac{e^{cos | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Let $I = \int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{(1+\cos^2 x)(e^{\cos x}+e^{-\cos x})} dx$  $(1)$Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:$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$$I = \int \limits_{0}^{\pi} \frac{e^{-\cos x} \sin x}{(1+\cos^2 x)(e^{-\cos x}+e^{\cos x})} dx$  $(2)$Adding $(1)$ and $(2)$:$2I = \int \limits_{0}^{\pi} \frac{(e^{\cos x} + e^{-\cos x}) \sin x}{(1+\cos^2 x)(e^{\cos x}+e^{-\cos x})} dx$$2I = \int \limits_{0}^{\pi} \frac{\sin x}{1+\cos^2 x} dx$Since $\frac{\sin x}{1+\cos^2 x}$ is symmetric about $x = \frac{\pi}{2}$,$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$Let $\cos x = t$,then $-\sin x dx = dt$. When $x=0, t=1$; when $x=\pi/2, t=0$.$I = -\int \limits_{1}^{0} \frac{dt}{1+t^2} = \int \limits_{0}^{1} \frac{dt}{1+t^2} = [	an^{-1} t]_{0}^{1} = 	an^{-1}(1) - 	an^{-1}(0) = \frac{\pi}{4}$

The value of $\int \limits_{0}^{\pi} \frac{e^{\cos x} \sin x}{\left(1+\cos ^{2} x\right)\left(e^{\cos x}+e^{-\cos x}\right)} d x$ is equal to

Similar Questions

Let $f(x)$ be a function satisfying $f(x) + f(\pi - x) = \pi^2, \forall x \in R$. Then $\int_{0}^{\pi} f(x) \sin x \, dx$ is equal to $...........$.

If $\int \limits_0^1 \frac{1}{\left(5+2 x -2 x ^2\right)\left(1+ e ^{(2-4 x)}\right)} dx =\frac{1}{\alpha} \log _{ e }\left(\frac{\alpha+1}{\beta}\right)$ where $\alpha, \beta > 0$,then $\alpha^4-\beta^4$ is equal to:

Let $f(x) = \frac{\sin x}{x}$,then $\int_{0}^{\frac{\pi}{2}} f(x) f\left(\frac{\pi}{2} - x\right) dx =$

$\int\limits_0^{\frac{\pi }{2}} {\sqrt {\sin 2\theta } } \sin \theta \,d\theta$ is equal to :

$\int_{-\pi}^{\pi} x^2 \sin x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module