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(Question) Trigonometric substitution triangles

  1. Sep 2, 2015 #1
    When we do trigonometric substitutions (such as for the integral x^3/(a^2-x^2)^2), we say something like "let x = asinp for -pi/2 <= p <= pi/2" then we carry on and solve the integral.

    However, sometimes our answer is ugly and we get some term in our expression like "cosp"- so we draw a little right angle triangle (SOH CAH TOA) to find out that "cosp = sqrt(a^2-x^2)/a <==> p = arccos(sqrt(a^2-x^2))" We then substitute this into cosp to get cos(arccos(Q(x))) = Q(X). Now our answer is nice and it is expressed in terms of x again.

    [My Question]
    Do we have to place a domain restriction on our answer when we do this? Because to me it is pretty clear that the SOH-CAH-TOA triangle only gives us valid results for x>0 and p>0

    What I mean is that sinp = x/a and cosp = sqrt(a^2-x^2)/a is only true for p>=0

    But beforehand we already defined our domain to be -pi/2 <= p <= pi/2 (when we did our sine substitution). However, we got "cosp" in our answer, so we have to express it in terms of x using arccos (but as stated above, this is only true for p>0)

    So is it true that if we use "cosp = sqrt(a^2-x^2)" anywhere we now have to make our new domain p <= pi/2 ?

    In all of the trig substitution questions I've done in calculus related courses, they've never asked for the domain, it was more of a mindless drone thing to do- and I'm really curious about this question.
  2. jcsd
  3. Sep 3, 2015 #2

    Simon Bridge

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    You do need to pay attention to the triangke you draw and how you have defined the angles.
    Some diagrams should hrlp you see what is going on better.
  4. Sep 3, 2015 #3
    Right, but one thing kind of bugs me.

    Say we have x = sinp, defined on p in [-pi/2, pi/2]

    we then have p = arcsinx and p = arccos (sqrt(a^2-x^2) )

    So there is a weird loss of information here, because whatever we put into arccos it will always output something in [0, pi/2].

    So we essentially "lost" all of the information about [-pi/2, 0] when we define p = arccos (sqrt(a^2-x^2) )

    This is kind of weird, and it is bugging me. Doesnt this loss of information somehow affect our intregral when we do any kind of cosine substitution?
  5. Sep 3, 2015 #4


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    You can avoid these "domain restrictions" by using the trig identities related to the "right triangle" computations you are doing.
  6. Sep 10, 2015 #5

    Simon Bridge

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    Like I ssid, draw a diagram of what yoh are doing to see if you actually lose the information you suspect, and how to go about recovering it if you have.
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