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Thread: Indefinite Integral without Trig Substitution: integrate (dx/(sqrt(-4x - x^2)))

  1. #11
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    Quote Originally Posted by debased View Post
    No worries, I've been staring at this for hours and I can't figure out where I went wrong. Can you show me the steps you did to integrate?
    It would be more instructive for you to write out your work (which might lead to you seeing the problem yourself); but I understand the frustration.

    We'll start with INT du/[2 sqrt(1-(u/2)^2)], as I said.

    We substitute z = u/2, so that u = 2z and du = 2 dz. This gives INT 2 dz/[2 sqrt(1-z^2)].

    The 2's cancel, so this is INT dz/sqrt(1-z^2) = arcsin(z) = arcsin(u/2) = arcsin((x+2)/2) = arcsin((1/2)x + 1).

    So, what did you do differently?

  2. #12
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    Quote Originally Posted by Dr.Peterson View Post
    It would be more instructive for you to write out your work (which might lead to you seeing the problem yourself); but I understand the frustration.

    We'll start with INT du/[2 sqrt(1-(u/2)^2)], as I said.

    We substitute z = u/2, so that u = 2z and du = 2 dz. This gives INT 2 dz/[2 sqrt(1-z^2)].

    The 2's cancel, so this is INT dz/sqrt(1-z^2) = arcsin(z) = arcsin(u/2) = arcsin((x+2)/2) = arcsin((1/2)x + 1).

    So, what did you do differently?
    I didn't do the second substitution with z. So you have to get it in the exact form of the integral i.e. 1/(sqrt(1 - x^2))? Because I did the first substitution and reached

    (1/2) * integral of du/(sqrt(1 - (u/2)^2)) and saw the (u/2)^2 and thought that is good to go to fit as the arcsin. But I guess like the chain rule for differentiation you have to keep substituting until you get it in that exact form?
    Last edited by debased; 12-07-2018 at 07:40 PM.

  3. #13
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    Quote Originally Posted by debased View Post
    I didn't do the second substitution with z. So you have to get it in the exact form of the integral i.e. 1/(sqrt(1 - x^2))? Because I did the first substitution and reached

    (1/2) * integral of du/(sqrt(1 - (u/2)^2)) and saw the (u/2)^2 and thought that is good to go to fit as the arcsin. But I guess like the chain rule for differentiation you have to keep substituting until you get it in that exact form?
    Exactly. In fact, substitution is the chain rule in reverse. The reason we have to do substitution is to make the integral you are working on exactly match a known integral, not just look generally like it.

    Incidentally, many tables of integrals give a more general form than the one you presumably know. For example, see #16 and #17 here. What they have done here is to "pre-digest" the substitution for you in #17. (Both can be done by trig substitution, which is no different from substitution in general except that you are replacing a variable with an expression rather than an expression with a variable, and you have to learn to recognize which one will work.)

  4. #14
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    Quote Originally Posted by Dr.Peterson View Post
    Exactly. In fact, substitution is the chain rule in reverse. The reason we have to do substitution is to make the integral you are working on exactly match a known integral, not just look generally like it.

    Incidentally, many tables of integrals give a more general form than the one you presumably know. For example, see #16 and #17 here. What they have done here is to "pre-digest" the substitution for you in #17. (Both can be done by trig substitution, which is no different from substitution in general except that you are replacing a variable with an expression rather than an expression with a variable, and you have to learn to recognize which one will work.)
    Wow that seems rather important and my teacher didn't cover that. In fact, knowing that it's the chain rule in reverse makes this a lot easier to look at for me. Thanks for your help!

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