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Indefinite integral. Different results depending upon what substitutions you use.

  1. May 15, 2012 #1
    ∫(tanxsec[itex]^{2}x[/itex])dx <---Original function to integrate.

    I do:

    ∫(tanxsec[itex]^{2}x[/itex])dx ---> ∫(tanx[itex]\frac{1}{cos^{2}x}[/itex])dx ---> ∫(([itex]\frac{sinx}{cosx}[/itex])([itex]\frac{1}{cos^{2}x}[/itex]))dx ---> ∫([itex]\frac{sinx}{cos^{3}x}[/itex])dx ---> substitution: [t=cosx , dt=-sinxdx -> [itex]\frac{dt}{-sinx}[/itex]=dx] ---> ∫([itex]\frac{sinx}{t^{3}}[/itex])[itex]\frac{dt}{-sinx}[/itex] ---> -∫(t[itex]^{-3}[/itex])dt ---> -[itex]\frac{t^{-2}}{-2}[/itex] + k ---> [itex]\frac{1}{2t^{2}}[/itex] + k ---> [itex]\frac{1}{2cos^{2}x}[/itex] + k <---My answer


    However my book gives [itex]\frac{tan^{2}x}{2}[/itex] +k as an answer, I followed their process and understand what they did, I just want to know if my answer is also correct and what would happend if this was a definite integral instead of an indefinite one? would arriving at different results create problems when computing definite integrals?
     
  2. jcsd
  3. May 15, 2012 #2
    Hey guys, searching around the web I found an explanation from MIT for this very same integral, here it is:

    ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-c-mean-value-theorem-antiderivatives-and-differential-equations/session-38-integration-by-substitution/MIT18_01SCF10_ex38sol.pdf

    So we know both my answer and the book's are correct but I still want to know how these differences among answers would affect the computation of a definite integral?
     
  4. May 15, 2012 #3
    Your result is good. They were wrong.
     
  5. May 15, 2012 #4
    Actually, you and the text are right!!

    [itex]\frac{1}{2}\mathrm{tan}^2(\theta) = \frac{1}{2}\frac{\mathrm{sin}^2(\theta)}{\mathrm{cos}^2(\theta )} = \frac{1}{2}\frac{1 - \mathrm{cos}^2(\theta )}{\mathrm{cos}^2(\theta)} = \frac{1}{2}\frac{1}{\mathrm{cos}^2(\theta)}-\frac{1}{2}\frac{\mathrm{cos}^2(\theta)}{\mathrm{cos}^2(\theta)}[/itex]

    So just absorb the [itex]-\frac{1}{2}[/itex] into the constant.

    To answer your question about definite integral: As you know the arbitrary constants will cancel out. The above computation shows that both indefinite integrals yield the same definite integral (regardless of the constant you pick).
     
  6. May 15, 2012 #5
    Hehe, interesting. Thanks.
     
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