Question

Q. Integration of 1/(cosec x -1) with respect to x . It's urgent.

Answer 1

Answer:

$\int \frac{1}{cosec \: x - 1} dx \\ \\ = \int \frac{1}{ \frac{1}{sin \: x } - 1} dx \\ \\ = \int \frac{sin \: x}{1 - sin \: x} dx \\ \\= \int \frac{sin \: x}{(1 - sin \: x)} \times \frac{(1 + sin \: x)}{(1 + sin \: x)} dx \\ \\= \int \frac{sin \: x + {sin}^{2} \: x }{(1 - sin^{2} \: x)} dx \\ \\= \int \frac{sin \: x + {sin}^{2} \: x }{ cos^{2} \: x} dx \\ \\= \int( \frac{sin \: x}{cos^{2} \: x} + \frac{{sin}^{2} \: x }{ cos^{2} \: x} ) dx \\ \\= \int( sec \: x \: tan \: x+ tan^{2} \: x) dx \\ \\ = \int(sec \: x \: tan \: x+ sec^{2} \: x - 1) dx \\ \\= \int \: sec \: x \: tan \: x \: dx + \int \: sec^{2} \: x \: dx - \int \: 1 \: dx \\ \\ = sec \: x \: + tan \: x - x + c \\ \\ \implies \\ \\ \purple{{\int \frac{1}{cosec \: x - 1} \: dx }} \purple{{= sec \: x \: + tan \: x - x + c }}$