int_{ - pi /2}^{pi /2} {frac{{cos x}}{{1 + {e | vedclass

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

Question

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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}dx}$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:$I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos( - \pi /2 + \pi /2 - x)}}{{1 + {e^{ - \pi /2 + \pi /2 - x}}}}dx} = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos( - x)}}{{1 + {e^{ - x}}}}dx} = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + e^{ - x}}}}dx$.Since $\cos(-x) = \cos x$,we have $I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + 1/e^x}}}dx = \int_{ - \pi /2}^{\pi /2} {\frac{{e^x \cos x}}{{e^x + 1}}}dx$.Adding the two expressions for $I$:$2I = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + e^x}}}dx + \int_{ - \pi /2}^{\pi /2} {\frac{{e^x \cos x}}{{1 + e^x}}}dx = \int_{ - \pi /2}^{\pi /2} {\frac{{\cos x(1 + e^x)}}{{1 + e^x}}}dx = \int_{ - \pi /2}^{\pi /2} {\cos x} dx$.$2I = [\sin x]_{ - \pi /2}^{\pi /2} = \sin(\pi /2) - \sin( - \pi /2) = 1 - (-1) = 2$.Therefore,$I = 1$.

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

Similar Questions

If $\int_0^{2024 \pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x=k$,then $\left(\frac{2 k}{\pi}+1\right)=$

The function $F(x) = \int_0^x \log \left( \frac{1 - t}{1 + t} \right) \,dt$ is

The value of the integral $\int_{-1}^{1} \left\{ \frac{x^{2013}}{e^{|x|}(x^{2}+\cos x)} + \frac{1}{e^{|x|}} \right\} dx$ is equal to

$\int_{0}^{1} \sin \left( 2 \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right) \, dx = $

$\int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module