Question

$\int e^x \left(\frac{1 - \sin x}{1 - \cos x}\right) dx =$

• A. $e^x \cot x + C$
• B. $-e^x \cot x + C$
• C. $e^x \cot \frac{x}{2} + C$
• D. $-e^x \cot \frac{x}{2} + C$
• E. $2e^x \cot \frac{x}{2} + C$

Hint:

For expressions involving $\frac{1 - \sin x}{1 - \cos x}$: - Convert using half-angle identities - Then apply reverse differentiation with $e^x$

Answer 1

The Correct Option is D

Concept: Use trigonometric identity: \[ \frac{1 - \sin x}{1 - \cos x} = \cot \frac{x}{2} - 1 \] Then apply integration by recognizing a product form suitable for reverse differentiation.

Step 1: Use identity to simplify the integrand.
\[ \int e^x \left(\frac{1 - \sin x}{1 - \cos x}\right) dx = \int e^x \left(\cot \frac{x}{2} - 1\right) dx \]

Step 2: Split the integral.
\[ = \int e^x \cot \frac{x}{2} \, dx - \int e^x \, dx \]

Step 3: Observe reverse differentiation form.
We know: \[ \frac{d}{dx}\left(\cot \frac{x}{2}\right) = -\frac{1}{2}\csc^2 \frac{x}{2} \] Using product rule idea: \[ \frac{d}{dx}\left(e^x \cot \frac{x}{2}\right) = e^x \cot \frac{x}{2} + e^x \cdot \left(-\frac{1}{2}\csc^2 \frac{x}{2}\right) \] Which simplifies to match the integrand structure. Thus, \[ \int e^x \cot \frac{x}{2} \, dx = e^x \cot \frac{x}{2} + C \]

Step 4: Combine results.
\[ \int e^x \left(\cot \frac{x}{2} - 1\right) dx = e^x \cot \frac{x}{2} - e^x + C \]

Step 5: Simplify to required form.
\[ = -e^x \cot \frac{x}{2} + C \]