Question

The value of \( \int e^{\tan \theta} (\sec \theta - \sin \theta) \, d\theta \) is

• A. \( e^{\tan \theta} \sec \theta + c \)
• B. \( e^{\tan \theta} \sin \theta + c \)
• C. \( e^{\tan \theta} (\sec \theta + \sin \theta) + c \)
• D. \( e^{\tan \theta} \cos \theta + c \)

Hint:

If an integral looks like \( e^{\text{function}} \times (\text{something}) \), always try to differentiate the exponential's power and see if you can partition the remaining part into "derivative of something" and "the product of the derivative and that something."

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This integral follows the form \( \int e^{g(x)} [g'(x)f(x) + f'(x)] \, dx = e^{g(x)}f(x) + c \).

Step 2: Key Formula or Approach
Let \( g(\theta) = \tan \theta \), so \( g'(\theta) = \sec^2 \theta \). We need to rewrite the bracket \( (\sec \theta - \sin \theta) \) to fit the form \( \sec^2 \theta \cdot f(\theta) + f'(\theta) \).

Step 3: Detailed Explanation
Let \( f(\theta) = \cos \theta \). Then \( f'(\theta) = -\sin \theta \). Now check the expression: \[ g'(\theta)f(\theta) + f'(\theta) = (\sec^2 \theta)(\cos \theta) + (-\sin \theta) \] \[ = \left(\frac{1}{\cos^2 \theta}\right)(\cos \theta) - \sin \theta = \sec \theta - \sin \theta \] This perfectly matches the integrand. Therefore: \[ \int e^{\tan \theta} (\sec \theta - \sin \theta) \, d\theta = e^{\tan \theta} \cos \theta + c \]

Step 4: Final Answer
The result of the integration is \( e^{\tan \theta} \cos \theta + c \).