int_{0}^{pi} frac{e^{cos x}}{e^{cos x}+e^{-co | vedclass

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

Question

(D) ધારો કે $I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x \quad ...(1)$ ગુણધર્મ $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) ધારો કે $I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x \quad ...(1)$ ગુણધર્મ $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$ નો ઉપયોગ કરતા,આપણને મળે: $I = \int_{0}^{\pi} \frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)}+e^{-\cos(\pi-x)}} d x$ કારણ કે $\cos(\pi-x) = -\cos x$,તેથી: $I = \int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} d x \quad ...(2)$ સમીકરણ $(1)$ અને $(2)$ નો સરવાળો કરતા: $2I = \int_{0}^{\pi} \left( \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} + \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} \right) d x$ $2I = \int_{0}^{\pi} \frac{e^{\cos x} + e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} d x$ $2I = \int_{0}^{\pi} 1 d x$ $2I = [x]_{0}^{\pi} = \pi - 0 = \pi$ $I = \frac{\pi}{2}$

$\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x=$

Similar Questions

$\int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x=$

$\sum_{n=1}^{10} \int_{-2n-1}^{-2n} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$ ની કિંમત શોધો.

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{e^x(x \sin x)}{e^{2x}-1} dx =$

જો $f(x) = \int\limits_0^x {2(\cos^2 3t + 3\sin^2 3t)dt}$ હોય,તો $f(x + \pi)$ ની કિંમત શું થાય?

કિંમત શોધો: $\int_{-a}^a f(x) dx - \int_0^a f(-x) dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module