Question

If \( \int \frac{e^x (1 + \sin x)}{1 + \cos x} \, dx = e^x f(x) + C \), then \( f(x) \) is equal to:

• A. \( \sin \frac{x}{2} \)
• B. \( \cos \frac{x}{2} \)
• C. \( \tan \frac{x}{2} \)
• D. \( \log \frac{x}{2} \)

Hint:

For integrals involving trigonometric expressions, use half-angle identities: \[ 1 + \cos x = 2 \cos^2 \frac{x}{2}, \quad 1 + \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}. \] This simplifies many fraction-based trigonometric integrals.

Answer 1

The Correct Option is C

Step 1: Consider the given integral: \[ I = \int \frac{e^x (1 + \sin x)}{1 + \cos x} \, dx. \] Using the trigonometric identity: \[ 1 + \cos x = 2 \cos^2 \frac{x}{2}, \quad 1 + \sin x = 2 \cos \frac{x}{2} \sin \frac{x}{2}, \] we rewrite the integral as: \[ I = \int \frac{e^x \cdot 2 \cos \frac{x}{2} \sin \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \, dx. \] Step 2: Simplify the expression: \[ I = \int e^x \cdot \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \, dx. \] \[ I = \int e^x \tan \frac{x}{2} \, dx. \] Step 3: Comparing with the given integral form: \[ I = e^x f(x) + C. \] Thus, we identify: \[ f(x) = \tan \frac{x}{2}. \]