Find the integral: $\int \frac{1}{1 + \tan x} \, dx$

$\frac{x}{2} + \frac{1}{2} \ln |\sin x + \cos x| + C$

The correct answer is Option (2) → $\frac{x}{2} + \frac{1}{2} \ln |\sin x + \cos x| + C$

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

$= \frac{1}{2} \int \frac{(\cos x + \sin x + \cos x - \sin x) \, dx}{\cos x + \sin x}$

$= \frac{1}{2} \int dx + \frac{1}{2} \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx$

$= \frac{x}{2} + \frac{C_1}{2} + \frac{1}{2} \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx$

Now, consider $I = \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx$

Put $\cos x + \sin x = t$ so that $(\cos x - \sin x) \, dx = dt$

Therefore $I = \int \frac{dt}{t} = \log |t| + C_2 = \log |\cos x + \sin x| + C_2$

Putting it in (1), we get

$\int \frac{dx}{1 + \tan x} = \frac{x}{2} + \frac{C_1}{2} + \frac{1}{2} \log |\cos x + \sin x| + \frac{C_2}{2}$

$= \frac{x}{2} + \frac{1}{2} \log |\cos x + \sin x| + \frac{C_1}{2} + \frac{C_2}{2}$

$= \frac{x}{2} + \frac{1}{2} \log |\cos x + \sin x| + C, \left( C = \frac{C_1}{2} + \frac{C_2}{2} \right)$