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Two different answers for the same integral?

  1. Jan 17, 2016 #1
    1. The problem statement, all variables and given/known data
    The anti-derivative of ∫##\frac{sinx}{sin^2x+4cos^2x}## is ##\frac{1}{\sqrt{3}}tan^{-1}((\frac{1}{\sqrt{3}})g(x))+C## then ##g(x)## is equal to :

    a. ##secx##
    b. ##tanx##
    c. ##sinx##
    d. ##cosx##

    2. Relevant equations
    ##d(cosx)=-sinx dx##

    3. The attempt at a solution

    I tried the problem the following way:

    [tex]\int \frac{sinx}{sin^2x+4cos^2x}\, dx=\int \frac{sinx}{1+3cos^2x}\, dx[/tex]
    Let ##t=cosx##. Therefore##-dt=sinx dx##
    Therefore [tex]I=\int \frac{-dt}{1+3t^2}=-\frac{1}{\sqrt{3}}tan^{-1}(\sqrt{3}cosx)[/tex]

    But my text gives the solution the following way:

    [tex]\int \frac{sinx}{sin^2x+4cos^2x}\, dx=\int \frac{tanxsecx}{tan^2x+4}\, dx=\int \frac{tanxsecx}{sec^2x+3}\, dx[/tex]. Let ##t=secx##, therefore:

    [tex]I=\int \frac{dt}{t^2+3}=\frac{1}{\sqrt{3}}tan^{-1}(\frac{secx}{\sqrt{3}})[/tex]

    So
    My answer is : [tex]I_m=-\frac{1}{\sqrt{3}}tan^{-1}(\sqrt{3}cosx)+C[/tex]
    Answer in my textbook is: [tex]I_t=\frac{1}{\sqrt{3}}tan^{-1}(\frac{secx}{\sqrt{3}})+C[/tex]

    And, I can't find any way to convert ##I_m## to ##I_t##.

    Am I wrong somewhere?
     
  2. jcsd
  3. Jan 17, 2016 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    $$\tan^{-1}\left( \frac{1}{x} \right) = \left\{
    \begin{array}{lr}
    \frac{\pi}{2} - \tan^{-1}(x) : x>0\\
    -\frac{\pi}{2} - \tan^{-1}(x) : x<0
    \end{array}
    \right.
    $$
     
  4. Jan 17, 2016 #3
    Oh yes! That cleared it out. Thanks blue_leaf77.
     
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