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Integration (different substitution = different answer)

  1. Mar 24, 2007 #1
    Out of curiosity there are several trig functions that can be integrated (WITHOUT the use of trig identities) using Integration by Substitution.

    One particular example is this:
    sin(x)cos(x) dx

    Integrating this with substitution u = cos(x) works out fine.
    HOWEVER integrating with substitution u = sin(x) does not.

    Can someone please explain why this is, and how to distinguish what substitution to make?
  2. jcsd
  3. Mar 24, 2007 #2
    Why does the substitution u=sin(x) not work?
  4. Mar 24, 2007 #3

    Gib Z

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    In fact u=sin x actually works BETTER than u=cos x, no need to multiply by factors of negative 1 to get things to work.
  5. Mar 24, 2007 #4


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    your chosen example is a bad one,... but I can imagine that there are cases where only a "correct" substitution will greatly simplify things.. which is your objective in the first place. remember, integrate by substitution/change of variables works best when the integrand has both f(x) and its derviative present (as a product).
  6. Mar 24, 2007 #5
    You get different answers until you remember that there's a constant of integration. The answers differ by a constant (1/2), so they aren't really different.
  7. Mar 25, 2007 #6
    Sorry i meant u = cos(x) works fine but u = sin(x) doesn't. (My bad).

    Even if you didn't know about the different substitutions how would you know that you need to multiply by -1 to get the right answer (assuming you have no way to check your answer).

    I agree it may have been a bad example but this was a spur of the moment quesiton...in this integrand f(x) does have a derivative present thats why substitution is best here...this doesn't explain why the one substitution works.

    There is a constant of integration for everything you integrate, I am just concerned with the integrands solution. Are you saying that some substitutions don't work because they require extra info in getting the constant of integrations?

    I'm not trying to be arrogant, I just think there is a logical solution to it like everything else in math :smile:
  8. Mar 25, 2007 #7
    Let [tex]u = sin(x)[/tex].
    Then [tex]du = cos(x)dx[/tex], so
    [tex]\int \sin(x)\cos(x)dx = \int udu = \frac{1}{2}u^2 + C = \frac{1}{2}\sin^2(x) + C[/tex]

    Now instead, let [tex]u = cos(x)[/tex].
    Then [tex]du = -sin(x)dx[/tex], so
    [tex]\int \sin(x)\cos(x)dx = \int -udu = -\frac{1}{2}u^2 + C = -\frac{1}{2}\cos^2(x) + C[/tex]

    [tex]\frac{1}{2}\sin^2(x) = -\frac{1}{2}\cos^2(x) + \frac{1}{2}[/tex]
    so they differ by a constant. Where is there a problem?
  9. Mar 25, 2007 #8

    Gib Z

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    No I was talking about since the derivative of cos x is NEGATIVE sin x, but we only have a sin x in the integrand, the natural thing to do is to multiply everything inside the integral by -1, and the outside by -1, to even it out.
  10. Mar 26, 2007 #9
    Yes they do differ by a half.
    Thus, the two integrals are indeed different.

    My point is if you were doing an analytical problem, then your end result will differ (by a constant) depending on the substitution. So which result is right?

    Using MATLAB the integral of sin(x)cos(x) = -0.5[cos(x)]^2.
    So, the substitution u = sin(x) is incorrect.
    Hence certain substitutions matter? Why?
  11. Mar 26, 2007 #10
    What must you always include when doing an indefinite integral? An arbitrary constant of integration, and this is because the derivative of a constant function is zero. If you were doing a definite integral either substitution would give the exact same final answer because that constant ends up disappearing when it is added and subtracted.

    As a trivial example consider the function y=1, then both Y=x+1, and Z=x+3 are antiderivatives of y. Now suppose we want to evaluate this integral over the interval (1,2)

    so Z(2)-Z(1)=[2+3]-[1+3]=[2-1]+[3-3]=1

    And they evaluate to the same answer.
  12. Mar 26, 2007 #11
    I agree with you.

    Would it be fair to say that for an indefinite integral you may remove any sum/difference constants and that would be the solution? (regardless of the substitution).
  13. Mar 26, 2007 #12
    This seems to mostly be a case of you not understanding the result.

    The best way to think of it is that the MATLAB result is wrong, or rather incomplete (they are just taking the constant to be 0).

    It should say:

    [tex]\int \sin(x)\cos(x)dx = -\frac{1}{2}\cos^2(x) + C[/tex]

    Which is exactly the same as:

    [tex]\int \sin(x)\cos(x)dx = \frac{1}{2}\sin^2(x) + C_2[/tex]


    [tex]-\frac{1}{2}\cos^2(x) + C = \frac{1}{2}\sin^2(x) + C_2[/tex]

    You can always add an arbitrary constant to an indefinite integral. If you are worried about keeping track of constants, just "eat up" any new constants by creating a new one (in the example above one of the C's ate up the other C and a constant factor of 1/2).

    You need to understand the difference between a definite and an indefinite integral.
  14. Mar 27, 2007 #13
    You'll also see that when you do a definite integral, again, you'll get the same value. (try it with the example above.)
  15. Mar 27, 2007 #14
    Instead of doing it with a single example, you can prove that for all integrable functions, the constant of integration doesn't matter for definite integrals.

    [tex]\int f(x)\,dx = F(x) + C[/tex] implies that [tex]\int^b_a f(x)\,dx = ( F(b) + C) - ( F(a) + C ) = F(b) - F(a) + C - C = F(b) - F(a)[/tex].

    This means that both [tex]F(x)[/tex] and [tex]F(x) + C [/tex] give the same answer.

    Also, another quick way to see that both of your answers are right is just to differentiate them both and notice that they're equal.
    Last edited: Mar 27, 2007
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