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 Question:
We must evaluate a definite integral containing an exponential fractional argument inside a sine function.
Observe the limits of integration carefully: $\log(1/2)$ and $\log 2$.

Step 2: Key Formula or Approach:
Notice that $\log(1/2) = \log(2^{-1}) = -\log 2$.
The integral bounds are of the form $\int_{-a}^{a} f(x) \, dx$.
When faced with bounds from $-a$ to $a$, always check if the integrand $f(x)$ is an odd function or an even function.
If $f(-x) = -f(x)$, the function is odd, and the integral evaluates exactly to 0.

Step 3: Detailed Explanation:
Let the integrand be $f(x) = \sin \left( \frac{e^x - 1}{e^x + 1} \right)$.
We test for parity by substituting $-x$:
$$f(-x) = \sin \left( \frac{e^{-x} - 1}{e^{-x} + 1} \right)$$
To simplify the complex fraction, multiply the numerator and denominator of the inner argument by $e^x$:
$$f(-x) = \sin \left( \frac{e^{-x} \cdot e^x - 1 \cdot e^x}{e^{-x} \cdot e^x + 1 \cdot e^x} \right)$$
$$f(-x) = \sin \left( \frac{1 - e^x}{1 + e^x} \right)$$
Factor out a $-1$ from the numerator:
$$f(-x) = \sin \left( -\frac{e^x - 1}{e^x + 1} \right)$$
Since the sine function is an odd function ($\sin(-\theta) = -\sin(\theta)$):
$$f(-x) = -\sin \left( \frac{e^x - 1}{e^x + 1} \right) = -f(x)$$
Because $f(-x) = -f(x)$, the entire integrand is proven to be an odd function.
Therefore, integrating this odd function over the symmetric interval $[-\log 2, \log 2]$ yields zero.
$$\int_{-\log 2}^{\log 2} \sin \left( \frac{e^x - 1}{e^x + 1} \right) dx = 0$$

Step 4: Final Answer:
The integral evaluates to 0, matching option (a).