The value of int_0^1 frac{dx}{e^x + e^{-x}} i | vedclass

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

Question

(B) We have the integral $I = \int_0^1 \frac{dx}{e^x + e^{-x}}$.Multiply the numerator and denominator by $e^x$:$I = \int_0^1 \frac{e^x}{e^{2x} + 1} d

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{e - 1}{e + 1}ight)$

Vedclass · Answer

(B) We have the integral $I = \int_0^1 \frac{dx}{e^x + e^{-x}}$.Multiply the numerator and denominator by $e^x$:$I = \int_0^1 \frac{e^x}{e^{2x} + 1} dx$.Let $t = e^x$,then $dt = e^x dx$.When $x = 0$,$t = e^0 = 1$. When $x = 1$,$t = e^1 = e$.Substituting these into the integral:$I = \int_1^e \frac{dt}{1 + t^2} = [	an^{-1} t]_1^e$.Evaluating the limits:$I = 	an^{-1}(e) - 	an^{-1}(1) = 	an^{-1}(e) - \frac{\pi}{4}$.Using the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1}\left(\frac{x - y}{1 + xy}ight)$:$I = 	an^{-1}\left(\frac{e - 1}{1 + e \cdot 1}ight) = 	an^{-1}\left(\frac{e - 1}{e + 1}ight)$.

The value of $\int_0^1 \frac{dx}{e^x + e^{-x}}$ is

Similar Questions

The value of the definite integral $\int_0^1 \frac{x \, dx}{x^3 + 16}$ lies in the interval $[a, b]$. The smallest such interval is

$\int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,dx = } $

Evaluate the definite integral $\int_2^5 2[x] \, dx$,where $[x]$ denotes the greatest integer function $\leq x$.

The value of the integral $\int_0^4 \frac{d x}{1+x^2}$ obtained by using the Trapezoidal rule with $h=1$ is

Evaluate the definite integral: $\int_{0}^{1} \left(1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \cdots \infty\right) e^{2x} \, dx$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module