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  4. Integrate (1/1-cotx) or (sinx/sinx-cosx) .

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

IIT JEEIIT JEE 1978

Solution:

Let I $= \int \limits \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} $
$\therefore I = \int \limits \frac {\frac {1}{2}(cosx+sinx)+ \frac {1}{2}(sinx-cosx)}{(sinx-cosx)}dx$
$ =\frac {1}{2} \int \limits \frac {cosx+sinx}{sinx-cosx}dx + \frac {1}{2} \int \limits 1 dx + c $
$ = \frac {1}{2}log (sinx-cosx) + \frac {1}{2} x + c $