”Calculate
Answer 1
Daniel Carter
To calculate the integral:
$ \int_{-1}^{1} \frac{1}{x + \sqrt{x^2 – 1}} dx $
First, consider the substitution $ x = \cosh(t) $, which implies $ dx = \sinh(t) dt $.
When $ x = -1 $, $ t = i \pi $ and when $ x = 1 $, $ t = 0 $:
$ \int_{i \pi}^{0} \frac{1}{\cosh(t) + \sinh(t)} \sinh(t) dt $
Knowing that $ \cosh(t) + \sinh(t) = e^t $, the integral becomes:
$ \int_{i \pi}^{0} \frac{\sinh(t)}{e^t} dt = \int_{i \pi}^{0} e^{-t} dt $
Evaluating this gives:
$ [ -e^{-t} ]_{i \pi}^{0} = -e^{0} + e^{-i \pi} = -1 + (-1) = -2 $
Answer 2
Emma Johnson
To calculate the integral:
$ int_{-1}^{1} frac{1}{x + sqrt{x^2 – 1}} dx $
First, consider the substitution $ x = cosh(t) $, which implies $ dx = sinh(t) dt $.
When $ x = -1 $, $ t = i pi $ and when $ x = 1 $, $ t = 0 $:
$ int_{i pi}^{0} frac{1}{cosh(t) + sinh(t)} sinh(t) dt $
Knowing that $ cosh(t) + sinh(t) = e^t $, the integral becomes:
$ int_{i pi}^{0} frac{sinh(t)}{e^t} dt = int_{i pi}^{0} e^{-t} dt $
Evaluating this gives:
$ [ -e^{-t} ]_{i pi}^{0} = -1 – 1 = -2 $
Answer 3
Michael Moore
To calculate the integral:
$ int_{-1}^{1} frac{1}{x + sqrt{x^2 – 1}} dx $
Consider the substitution $ x = cosh(t) $:
$ int_{i pi}^{0} frac{sinh(t)}{e^t} dt $
Evaluating this gives:
$ -2 $
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