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: $$f(x) = \tan \frac{x}{2}.$$