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Math Help - Evaluate the Indefinite Integral

  1. #1
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    Evaluate the Indefinite Integral

    Hi,
    I am having trouble evaluating the indefinite integral of the following problem using U-substitution. The answer at the back of the book is 2sin square root of t + C. I'm not sure what U should be. Any assistance will be appreciated. Thanks in advance.

    1. The problem statement, all variables and given/known data

    Evaluate the indefinite integral

    2. Relevant equations

    integral of cos * (square root of t) / (square root of t) dt

    3. The attempt at a solution

    integral of cos * (square root of t) / (square root of t) dt

    Let U = ?

    Lets say U = cos * (square root of t)

    du = sin * (square root of t) dx

    This does not make any sense to me. Im I going right? Where did I go wrong?
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  2. #2
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    Edit: Nevermind, that's derivatives. Hold on.
    Last edited by ZevWolf; April 6th 2010 at 04:04 PM.
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  3. #3
    Super Member Deadstar's Avatar
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    \cos(\sqrt{t}) = 1 - \frac{(\sqrt{t})^2}{2!} + \frac{(\sqrt{t})^4}{4!} - \frac{(\sqrt{t})^6}{6!} + ...

    So,

    \frac{\cos(\sqrt{t})}{\sqrt{t}} = \frac{1}{\sqrt{t}} - \frac{(\sqrt{t})^1}{2!} +  \frac{(\sqrt{t})^3}{4!} - \frac{(\sqrt{t})^5}{6!} + ...

    = \sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{t})^{2n-1}}{(2n)!}

    So the integral of this is...

    \sum_{n=0}^{\infty} \frac{2t(-1)^n (\sqrt{t})^{2n-1}}{(2n+1) (2n)!} + C = \sum_{n=0}^{\infty} \frac{2(-1)^n (\sqrt{t})^{2n+1}}{(2n+1)!} + C.

    Now the power series of \sin(\sqrt{t}) is...

    \sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{t})^{2n+1}}{(2n+1)!}...

    So... Can ya finish..?
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  4. #4
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    Thanks for taking the time to help me with this Deadstar. I really appreciate this.

    Wow, what you posted was an eyefull, unfortunately, I don't understand. Ok, I am new to this, but from my understanding (correct me if I am wrong), I am supposed to use U-substitution, meaning I am supposed to assign U = something, then find du = the derivative of something dx, etc etc. Can you tell me what should I assign U = to?
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  5. #5
    Super Member Deadstar's Avatar
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    Quote Originally Posted by sparky View Post
    Thanks for taking the time to help me with this Deadstar. I really appreciate this.

    Wow, what you posted was an eyefull, unfortunately, I don't understand. Ok, I am new to this, but from my understanding (correct me if I am wrong), I am supposed to use U-substitution, meaning I am supposed to assign U = something, then find du = the derivative of something dx, etc etc. Can you tell me what should I assign U = to?
    Do you have to solve it that way or is that what you think you have to do. I did not use a u-substitution method.

    The way I solved it was to expand \cos(\sqrt{t}) as a power series.

    See part 3 here. Trigonometric functions - Wikipedia, the free encyclopedia

    Then divided it through by \sqrt{t} to get the integral in your question.

    Then I used the fact...

    The integral of a sum is the sum of an intergal.

    This basically just follows from the following property of integrals.

    \int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx

    So all I had to do was integrate the sum formula that generated my series which was...

    = \sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{t})^{2n-1}}{(2n)!}

    Which is equal to the power series for \cos(x) multiplied by \frac{1}{\sqrt{t}}.

    I.e. I did this...

    = \int \sum_{n=0}^{\infty} \frac{(-1)^n  (\sqrt{t})^{2n-1}}{(2n)!} dt

    = \sum_{n=0}^{\infty} \int \frac{(-1)^n  (\sqrt{t})^{2n-1}}{(2n)!} dt

    = \sum_{n=0}^{\infty} \frac{2t(-1)^n (\sqrt{t})^{2n-1}}{(2n+1) (2n)!} + C

    (why?)

    because...

    \sum_{n=0}^{\infty} \int \frac{(-1)^n  (\sqrt{t})^{2n-1}}{(2n)!} dt

    = \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \int (\sqrt{t})^{2n-1} dt

    = \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \int t^{1/2(2n-1)} dt

    = \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \int t^{n-1/2} dt

    = \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \frac{t^{n+1/2}}{n+1/2} dt

    = \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \frac{2t^{n+1/2}}{2n+1} dt

    = \sum_{n=0}^{\infty} \frac{2(-1)^n (\sqrt{t})^{2n+1}}{(2n+1)!} + C

    Then I just realized it was almost the same as the power series for \sin(\sqrt{t}) but multiplied by 2.

    Have a look at that link and you'll see the series expansions for cos and sin. Just put \sqrt{t} where the x's are and you get my series.
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  6. #6
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    hi

    wow!!!!!!!! its nice but sparky your right with your idea...he just explore it to understand be easily...keep doin that i like calculus so much actually all kinds of math...









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