# Integration of Cosecx

1. Dec 1, 2013

### kashan123999

1. The problem statement, all variables and given/known data

So I got towards the intergration of cosecx as -ln(cosecx + cotx),but in my coursebook it is ln(cosecx-cotx)?so how do you put that sign in the log :D

2. Relevant equations

∫[f(x)]/[F(x)] dx = ln|F(x)|+ c

3. The attempt at a solution

∫(cosecx) X (cosecx + cotx)/(cosecx + cotx) dx

∫[(cosecx)^2 + (cosecxcotx)]/(cosecx + cotx) dx

let cosecx + cot x = u → [-cosecxcotx - (cosecx)^2]dx = du

-[cosecxcotx + (cosecx)^2]dx = du

dx = -du/[cosecxcotx + (cosecx)^2]

so putting value of dx and (cosecx + cotx)

∫[(cosecxcotx) + (cosecx)^2] X (-du) / [(u) X {(cosecxcotx+ (cosecx)^2)}]

cosecxcotx + (cosecx)^2 will simply each other in numerator and denominator hence

∫(-du)/(u)

using reciprocal rule

-ln|u| + c

-ln|cosecx + cotx| + c

so in my courebook it is ln|cosecx - cotx|?? how to do that?

2. Dec 1, 2013

### Saitama

You can use this property of logarithm: $\log a^b=b\log a$

Therefore,
$$-\ln|\csc x+\cot x|=\ln\frac{1}{|\csc x+\cot x|}$$
Multiply and divide by $\csc x-\cot x$ to get the same answer as book.