Question

Integrate $\frac{1}{1-cotx}$ or $\frac{sinx}{sinx-cosx}.$

Answer 1

Let I $=\int \frac{sinx}{sinx-cosx}dx$
Again, let sinx=A(cosx+sinx)+B(sinx-cosx),
then A+B=1 and A-B=0
$\Rightarrow A=\frac{1}{2},B=\frac{1}{2}$
$\therefore I=\int \frac{\frac{1}{2}(cosx+sinx)+\frac{1}{2}(sinx-cosx)}{(sinx-cosx)}dx$
$=\frac{1}{2}\int \frac{cosx+sinx}{sinx-cosx}dx+\frac{1}{2}\int 1dx+c$
$=\frac{1}{2}log(sinx-cosx)+\frac{1}{2}x+c$