Question

Integrate the following function: $\frac{1}{1+ tan\ x}$

Answer 1

Solution:
$\int\frac{1}{1+ tan\ x} dx\\ \\ =\int \frac{ {sec}^{2}x }{ {sec}^{2} x(1 + tan \: x)} dx\\ \\\int \frac{ {sec}^{2}x }{ (1 + {tan}^{2} x)(1 + tan \: x)}dx \\ \\ put \: tan \: x = t \\ \\ {sec}^{2} x \: dx = dt \\ \\ = \int\frac{dt}{(1 + {t}^{2} )(1 + t)}$
This can be written as

$=\int\frac{1}{2(1 + t)} - \frac{1(t - 1)}{2( {t}^{2} + 1) } \\ \\ = \frac{1}{2} \int\frac{1}{1 + t}dt - \frac{1}{2} \int\frac{t}{1 + {t}^{2} }dt + \frac{1}{2} \int\frac{1}{1 + {t}^{2} }dt \\ \\ = \frac{1}{2} log |1 + t| - \frac{1}{4} log |1 + {t}^{2} | + \frac{1}{2} {tan}^{ - 1} (1 + {t}^{2} ) + C\\ \\ undo \: substitution\\ \\\int\frac{1}{1+ tan\ x} dx = \frac{1}{2} log |1 + tan \: x| - \frac{1}{4} log |1 + {tan}^{2}x | + \frac{1}{2} {tan}^{ - 1} (1 + {tan}^{2} x) + C$