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Math Help - 1 Quick integration problem

  1. #1
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    1 Quick integration problem

    Integral of (1/sqrt(x)) * cos((pi*sqrt(x))/2) in the range of 1 to 4. I'm supposed to integrate by substitution using u = pi*sqrt(x) all divided by 2

    Thus,

    dy = (pi*x^3/2)/3

    And I get

    integral of:

    3/pi * cos(u) du from the range of 1 to 4, but this does not take care of the problem with the 1/sqrt(x) in the original and the new x^(3/2) in the new one.
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  2. #2
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    Quote Originally Posted by Lord Darkin View Post
    Integral of (1/sqrt(x)) * cos((pi*sqrt(x))/2) in the range of 1 to 4. I'm supposed to integrate by substitution using u = pi*sqrt(x) all divided by 2

    Thus,

    dy = (pi*x^3/2)/3 what is this ?

    And I get

    integral of:

    3/pi * cos(u) du from the range of 1 to 4, but this does not take care of the problem with the 1/sqrt(x) in the original and the new x^(3/2) in the new one.
    \int_1^4 \frac{1}{\sqrt{x}} \cdot \cos\left(\frac{\pi \sqrt{x}}{2}\right) \, dx

    u = \frac{\pi \sqrt{x}}{2}

    du = \frac{\pi}{4\sqrt{x}} \, dx


    \frac{4}{\pi} \int_1^4 \frac{\pi}{4\sqrt{x}} \cdot \cos\left(\frac{\pi \sqrt{x}}{2}\right) \, dx<br />

    substitute and reset the limits of integration ...

    \frac{4}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos{u} \, du

    finish.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \int_1^4 \frac{1}{\sqrt{x}} \cdot \cos\left(\frac{\pi \sqrt{x}}{2}\right) \, dx

    u = \frac{\pi \sqrt{x}}{2}

    du = \frac{\pi}{4\sqrt{x}} \, dx


    \frac{4}{\pi} \int_1^4 \frac{\pi}{4\sqrt{x}} \cdot \cos\left(\frac{\pi \sqrt{x}}{2}\right) \, dx<br />

    substitute and reset the limits of integration ...

    \frac{4}{\pi} \int_{\frac{\pi}{2}}^{\pi} \cos{u} \, du

    finish.

    Ohhh my du was wrong. You got the right answer according to my answer key. Thanks. I forgot that I had to take the derivative of u, not take the integrand.

    Last question - would this still work if I resetted th limits to be 0 and (-1/2)? Cos(pi) = -1/2 and cos(pi/2) = 0
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  4. #4
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    Quote Originally Posted by Lord Darkin View Post
    Ohhh my du was wrong. You got the right answer according to my answer key. Thanks. I forgot that I had to take the derivative of u, not take the integrand.

    Last question - would this still work if I resetted th limits to be 0 and (-1/2)? Cos(pi) = -1/2 and cos(pi/2) = 0
    no, the limits of integration are related to the variable of integration which is u in this case.

    the antiderivative of \cos{u} is \sin{u} , evaluated from \frac{\pi}{2} to \pi using the Fundamental Theorem of Calculus.

    \sin(\pi) - \sin\left(\frac{\pi}{2}\right) = 0 - 1 = -1

    therefore, the value of the original definite integral is -\frac{4}{\pi}

    you need to review/relearn the process of finding the values of definite integrals by substitution.
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