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

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

    So I'm trying to understand how to integrate: (dx/(sqrt(-4x - x^2))) without using trig substitution as I'm in a Calc 1 class and we're not supposed to use anything other than u-substitution. I get stuck after completing the square:

    indefinite integral of (dx/(sqrt(-4x - x^2)))
    indefinite integral of (dx/(sqrt(-(x + 2)^2 + 4)))

    Let u = x + 2

    indefinite integral of (dx/(sqrt(4 - u^2)))

    Perhaps this is an algebra question, but I'm not sure how you get that 4 to be a 1 as it's inside a radical and part of a sum. What do I do from here?

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    Quote Originally Posted by debased View Post
    So I'm trying to understand how to integrate: (dx/(sqrt(-4x - x^2))) without using trig substitution as I'm in a Calc 1 class and we're not supposed to use anything other than u-substitution. I get stuck after completing the square:

    indefinite integral of (dx/(sqrt(-4x - x^2)))
    indefinite integral of (dx/(sqrt(-(x + 2)^2 + 4)))

    Let u = x + 2

    indefinite integral of (dx/(sqrt(4 - u^2)))

    Perhaps this is an algebra question, but I'm not sure how you get that 4 to be a 1 as it's inside a radical and part of a sum. What do I do from here?
    Until you integrate or differentiate any questions you have will be either algebra or arithmetic.

    Note that 4 =2^2.
    4-u^2 = 4(1 - u^2/4) = 4(1-(u/2)^2). So sqrt(4-u^2) = sqrt(4(1-(u/2)^2)) =sqrt(4)*sqrt((1-(u/2)^2) = 2*sqrt((1-(u/2)^2).

    Now what???
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

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    Quote Originally Posted by debased View Post
    So I'm trying to understand how to integrate: (dx/(sqrt(-4x - x^2))) without using trig substitution as I'm in a Calc 1 class and we're not supposed to use anything other than u-substitution. I get stuck after completing the square:

    indefinite integral of (dx/(sqrt(-4x - x^2)))
    indefinite integral of (dx/(sqrt(-(x + 2)^2 + 4)))

    Let u = x + 2

    indefinite integral of (dx/(sqrt(4 - u^2)))

    Perhaps this is an algebra question, but I'm not sure how you get that 4 to be a 1 as it's inside a radical and part of a sum. What do I do from here?
    You say you can only use u-substitution. I would call a trig substitution a kind of u-substitution, but I won't quibble over that.

    The question is, what basic forms have you learned, that you can handle after you substitute? Are you implying that you have been given a formula for the integral of, say, du/sqrt(1 - u^2)?

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    Quote Originally Posted by Jomo View Post
    Until you integrate or differentiate any questions you have will be either algebra or arithmetic.

    Note that 4 =2^2.
    4-u^2 = 4(1 - u^2/4) = 4(1-(u/2)^2). So sqrt(4-u^2) = sqrt(4(1-(u/2)^2)) =sqrt(4)*sqrt((1-(u/2)^2) = 2*sqrt((1-(u/2)^2).

    Now what???
    Then I get to 1/2 * arcsin(1/2x + 1) + C which doesn't work.

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    Quote Originally Posted by debased View Post
    Then I get to 1/2 * arcsin(1/2x + 1) + C which doesn't work.
    I presume you mean arcsin((1/2)x + 1).

    So you started with INT du/[2 sqrt(1-(u/2)^2)], and ended up with (1/2) arcsin((1/2)x + 1) + C. That's what I got, though I wrote it differently.

    What do you mean by saying it doesn't work? Do you mean you took the derivative and didn't end up with the integrand?

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    Quote Originally Posted by Dr.Peterson View Post
    I presume you mean arcsin((1/2)x + 1).

    So you started with INT du/[2 sqrt(1-(u/2)^2)], and ended up with (1/2) arcsin((1/2)x + 1) + C. That's what I got, though I wrote it differently.

    What do you mean by saying it doesn't work? Do you mean you took the derivative and didn't end up with the integrand?
    Yeah if I differentiate that I get 1/(2 sqrt(-x (4 + x)))

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    Quote Originally Posted by debased View Post
    Yeah if I differentiate that I get 1/(2 sqrt(-x (4 + x)))
    I don't.

    Please write out the details and check your work.

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    Quote Originally Posted by Dr.Peterson View Post
    I don't.

    Please write out the details and check your work.
    d/dx [ (1/2) * arcsin((1/2)x + 1) ]

    (1/2) * d/dx [ arcsin((1/2)x + 1) ]

    (1/2) * ( 1/(sqrt(1 - ((x+2)/2)^2)) * (1/2) )

    (1/4) * (1/(sqrt(1 - ((x+2)/2)^2)))

    1/(4 * sqrt(1 - ((x+2)/2)^2))

    1/(4 * sqrt((-x^2 - 4x)/4))

    1/(2 * sqrt(-4x - x^2))
    Last edited by debased; 12-07-2018 at 05:37 PM.

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    Quote Originally Posted by debased View Post
    d/dx [ (1/2) * arcsin((1/2)x + 1) ]

    (1/2) * d/dx [ arcsin((1/2)x + 1) ]

    (1/2) * ( 1/(sqrt(1 - ((x+2)/2)^2)) * (1/2) )

    (1/4) * (1/(sqrt(1 - ((x+2)/2)^2)))

    1/(4 * sqrt(1 - ((x+2)/2)^2))

    1/(4 * sqrt((-x^2 - 4x)/4))

    1/(2 * sqrt(-4x - x^2))
    Sorry about that -- there are too many 1/2's floating around. The result I got for the integral didn't have the 1/2 on the outside; it was just arcsin((1/2)x + 1).

    So we'll have to look at the last steps of your integration, rather than your check.

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    Quote Originally Posted by Dr.Peterson View Post
    Sorry about that -- there are too many 1/2's floating around. The result I got for the integral didn't have the 1/2 on the outside; it was just arcsin((1/2)x + 1).

    So we'll have to look at the last steps of your integration, rather than your check.
    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?

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