int_{frac{-pi}{2}}^{frac{pi}{2}} frac{cos x}{ | vedclass

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

Question

(A) Let $I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} dx$ ... $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} dx$ ... $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:$I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos(\frac{-\pi}{2} + \frac{\pi}{2} - x)}{1+e^{(\frac{-\pi}{2} + \frac{\pi}{2} - x)}} dx$$I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos(-x)}{1+e^{-x}} dx = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+\frac{1}{e^x}} dx$$I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{e^x \cos x}{e^x+1} dx$ ... $(2)$Adding equations $(1)$ and $(2)$:$2I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x + e^x \cos x}{1+e^x} dx$$2I = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x(1+e^x)}{1+e^x} dx = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x dx$Since $\cos x$ is an even function,$\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx$:$2I = 2\int_{0}^{\frac{\pi}{2}} \cos x dx$$I = \int_{0}^{\frac{\pi}{2}} \cos x dx = [\sin x]_{0}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1$.

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x=$

Similar Questions

Suppose $f$ is such that $f(-x) = -f(x)$ for every real $x$ and $\int_{0}^{1} f(x) dx = 5$,then $\int_{-1}^{0} f(t) dt = $

The value of $\int_0^{2\pi } {{\cos }^{99}x\,dx} $ is

If $f(5-x)=f(x)$ and $\int_2^3 f(x) dx=2$,then $\int_2^3 x f(x) dx=$

The value of $\int_0^{n\pi + v} {|\sin x|\,dx} $ is

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module