Question

\(\int_0^{\pi/2} \cos x \cdot e^{\sin x} \, dx\) is equal to

• A. \(e + 1\)
• B. \(e - 1\)
• C. \(e\)
• D. \(-e\)

Hint:

Recall $e^0 = 1$. The result is $e^1 - e^0 = e - 1$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Definite integration using substitution (u-substitution) and transformation of boundary limits.
Key Formula or Approach:
\[ \int f'(x) e^{f(x)} \, dx = e^{f(x)} + C \]

Step 2: Detailed Explanation:
Given the integral:
\[ I = \int_0^{\pi/2} \cos x \cdot e^{\sin x} \, dx \]
Let \(u = \sin x\).
Then \(du = \cos x \, dx\).
Transforming the limits of integration:
- Lower limit: At \(x = 0 \implies u = \sin(0) = 0\).
- Upper limit: At \(x = \frac{\pi}{2} \implies u = \sin\left(\frac{\pi}{2}\right) = 1\).
The transformed integral becomes:
\[ I = \int_0^1 e^u \, du = [e^u]_0^1 = e^1 - e^0 = e - 1 \]

Step 3: Final Answer:
Therefore, the integral is equal to \(e - 1\), matching option (B).