Question

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

Answer 1

$Let \ I = ∫\frac {1}{1-tan \ x } dx$

$I= ∫\frac {1}{1-\frac {sin\ x}{cos \ x }}dx$

$I= ∫\frac {cos\ x}{cos \ x - sin\ x} dx$

$I= \frac 12 ∫\frac {2cos\ x}{cos\ x - sin\ x} dx$

$I= \frac 12 ∫\frac {(cos \ x-sin\ x)+(cos\ x+sin \ x)}{(cos \ x-sin\ x) }dx$

$I= \frac 12 ∫1dx+\frac 12 ∫\frac {cos \ x+sin\ x}{cos\ x-sin\ x} dx$

$I= \frac x2+\frac 12 ∫\frac {cos \ x+sin\ x}{cos\ x-sin\ x} dx$

$Put \ cos\ x - sin\ x = t ⇒ (-sin\ x - cos \ x) dx = dt$

$∴ I = \frac x2+\frac 12 ∫\frac {-(dt)}{t}$

$I= \frac x2-\frac 12 log\ |t|+C$

$I= \frac x2-\frac 12 log\ |cos\ x-sin\ x|+C$