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Math Help - Tricky Substitutions for Integrals: Practice Problems

  1. #1
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    Tricky Substitutions for Integrals: Practice Problems

    These are nice problems that can enlighten a few people who are not used to non-routine substitutions. People who are well acquainted with this, can take one integral at a time and state the substitution and show the solution step by step.



    State the substitution and show the solution for the following integrals:

    1) Prove \int_a^b \sqrt{(x-a)(x-b)}\, dx = \frac{\pi (b-a)^2}8

    2) For n \in \mathbb{Z}^+ , a > |b|,.Prove that  \int_0^{\pi} (a + b\cos x)^{-n} \, dx = (a^2 - b^2)^{-(n - \frac12)} \int_0^{\pi} (a - b\cos y)^{n-1} \, dy. Evaluate the integral for n = 1,2,3.

    3) For m, n \in \mathbb{Z}^+. Prove \int_a^b (x-a)^m(b-x)^n \, dx = (b-a)^{m+n+1}\frac{m! n!}{(m+n+1)!}


    The aim of this thread was to let beginner-integral-lovers enjoy and learn something new. So feel free to skip a few obvious steps while writing the solution.

    P.S: I think I will not be available from tomorrow to the end of this month
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Isomorphism View Post
    1) Prove \int_a^b \sqrt{(x-a)(x-b)}\, dx = \frac{\pi (b-a)^2}8
    Huh ok, I'll do what you want :
    Code:
    SUBSTITUTION
    Result is pi (b-a)/8
    Am I right ?

    By the way, I had a question for this : if x is between a & b, then x-a>0 and x-b<0 (in the case b>a).
    So what's going on ?
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  3. #3
    Eater of Worlds
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    I am going to have to wait until Latex is fixed. I have tried Moo's link, but I can't get it to work for me.

    Anyway, here is the first one.

    I used an initial sub of u=(2x-a-b)/2, du=dx

    That gives:

    1/2INT[(2u-a+b)(2u+a-b)]du

    Now, I left u=((a-b)/2)sec(t), du=((a-b)/2)sec(t)tan(t)dt

    This results in:

    ((a-b)^2)/4*INT[tan^2(t)sec(t)]dt, t=0..Pi

    And the rest follows.
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by Moo View Post
    By the way, I had a question for this : if x is between a & b, then x-a>0 and x-b<0 (in the case b>a).
    So what's going on ?
    I don't think the integral exists in the real numbers set.
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  5. #5
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    I have an idea, but I'm not good at double integrals, so I'll need some help

    here it is:

    we define g(x,y) = x^2 - (a+b)x +ab - y^2

    now the problem is how to set the limits

    here's my suggestion: for x: a to b ; for y: 0 to the f_max

    now, f_max: f'(x) = 0 ;

    2x = a+b and x = (a+b)/2

    f_max = f((a+b)/2) = sqrt((2a-a^2-b^2)/2)

    Now we have our double Integral, which we're going to calculate using polar coordinates (where the pi should come from)

    what do you think, could someone post this solution, if it's correct and if not, please tell me where my mistake is
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