Question

$\int \frac{x + \sin x}{1 + \cos x} dx =$}

• A. $x \cos x + c$
• B. $x \tan x + c$
• C. $x \tan \frac{x}{2} + c$
• D. $x \sec^2 \frac{x}{2} + c$

Hint:

Look for the form $\int [f(x) + xf'(x)]dx = xf(x) + c$. Here $f(x) = \tan(x/2)$.

Answer 1

The Correct Option is C

Step 1: Concept
Use half-angle trigonometric identities: $1 + \cos x = 2 \cos^2(x/2)$ and $\sin x = 2 \sin(x/2) \cos(x/2)$.

Step 2: Meaning
$\int \frac{x}{2\cos^2(x/2)} dx + \int \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} dx = \frac{1}{2} \int x \sec^2(x/2) dx + \int \tan(x/2) dx$.

Step 3: Analysis
Apply integration by parts to $\frac{1}{2} \int x \sec^2(x/2) dx$. Let $u=x$, $dv = \frac{1}{2}\sec^2(x/2)dx \implies v = \tan(x/2)$. Integral $= x \tan(x/2) - \int \tan(x/2) dx$.

Step 4: Conclusion
The term $\int \tan(x/2) dx$ cancels out, leaving $x \tan(x/2) + c$. Final Answer: (C)