Question

$\int_{\log(1/2)}^{\log 2} \sin \left( \frac{e^x - 1}{e^x + 1} \right) dx = \_\_\_\_\_\_.$

• A. 0
• B. 1
• C. cos(1/2)
• D. 2 \log(1/2)

Hint:

Any algebraic term of the specific form $\frac{e^x - 1}{e^x + 1}$ or $\frac{a^x - 1}{a^x + 1}$ is intrinsically an odd function. Whenever you see these inside a symmetric integral, the answer is almost always 0!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Check if the function is odd or even over a symmetric interval $[-a, a]$. Note that $\log(1/2) = -\log 2$.

Step 2: Formula Application:
Let $f(x) = \sin\left(\frac{e^x - 1}{e^x + 1}\right)$. $f(-x) = \sin\left(\frac{e^{-x} - 1}{e^{-x} + 1}\right) = \sin\left(\frac{1/e^x - 1}{1/e^x + 1}\right)$.

Step 3: Explanation:
$f(-x) = \sin\left(\frac{1 - e^x}{1 + e^x}\right) = \sin\left(-\frac{e^x - 1}{e^x + 1}\right)$. Since $\sin(-\theta) = -\sin \theta$, we have $f(-x) = -f(x)$. The function is odd. The integral of an odd function over a symmetric interval $[-\log 2, \log 2]$ is always 0.

Step 4: Final Answer:
The integral equals 0.