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The Same Integral Two Different Answers

  1. Jun 10, 2007 #1
    The Same Integral... Two Different Answers!!!

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

    [tex] \int \frac{\tan x}{\cos^5 x} \,dx [/tex]

    3. The attempt at a solution

    Solution 1
    [tex] \int \frac{\tan x}{\cos^5 x} \,dx = \int \frac{\sin x}{\cos^6 x} [/tex]

    [tex]u = \cos x \,\,\,\, du = -\sin x \,dx [/tex]

    [tex]= -\int u^-^5 \,du = \frac{1}{4 u^4} + C [/tex]

    Answer 1: [tex] \frac{1}{4 \cos^4 x} + C [/tex]

    Solution 2

    [tex] \int \frac{\tan x}{\cos^5 x} \,dx = \int \tan x \sec^4 x \,dx = \int \tan x (1 + \tan^2 x) \sec^2 x \,dx [/tex]

    [tex]u = \tan x \,\,\,\, du = \sec^2 x \,dx [/tex]

    [tex] = \int \(u + u^3) \,du = \frac{1}{2} u^2 + \frac{1}{4} u^4 + C [/tex]

    Answer 2: [tex] \frac{1}{2} \tan^2 x + \frac{1}{4} \tan^4^ x + C[/tex]

    Wich one is right and why?
    Last edited: Jun 10, 2007
  2. jcsd
  3. Jun 10, 2007 #2
    In the first step of your second solution, you should have [tex]\sec^5{x}[/tex], not [tex]\sec^4{x}[/tex]. You're also off by a power in your first solution. Close, though.

    P.S.: Check your answers by taking derivatives and see if you arrive back at the integrand.
  4. Jun 10, 2007 #3


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    Gold Member

    In the first attempt you should have [tex] -\int u^-^6 \,du [/tex]

    And similarly in the second attempt you have replaced [tex]\frac{1}{cos^5(x)} [/tex] with sec to the power 4 which is not correct.
  5. Jun 10, 2007 #4
    I will tell you that neither of the answers are right. You find where the erroneous step is.(It's quite simple, really)
    [EDIT]Hmm...never mind.

    A further simplification of the second method is to write down the integral as
    [tex] \int \frac{\tan x}{\cos^5 x} \,dx = \int \tan x \sec^5 x \,dx = \int \tan x \sec x \sec^4 x \,dx [/tex]
  6. Jun 10, 2007 #5

    Gib Z

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    Homework Helper

    Whenever you seem to get two solutions, differentiate both of them. If they both arrive back at the integrand, then try and see why the 2 solutions have a difference of a constant, thats what the +C's there for = ]
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