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Homework Help: Integration of Cosecx

  1. Dec 1, 2013 #1
    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. jcsd
  3. Dec 1, 2013 #2
    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.
     
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