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Math Help - Integration - First Steps

  1. #1
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    Integration - First Steps

    Could anyone please help me with these integration problems? You don't have to give me the answers, but if you could point me in the direction to get started, I'd really appreciate it.

    Thanks in advance!

    1. \int\frac{1}{\sqrt{25-7t^2}}~dt

    2. \int\frac{x^3}{x^2-2x+3}~dx

    3. \int e^{3x}cos{4x}~dx
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  2. #2
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    2. \int\frac{x^3}{x^2-2x+3}~dx= \int x+2+\frac{x-6}{x^2-2x+3}~dx
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  3. #3
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    Quote Originally Posted by greenstupor View Post
    Could anyone please help me with these integration problems? You don't have to give me the answers, but if you could point me in the direction to get started, I'd really appreciate it.

    Thanks in advance!

    1. \int\frac{1}{\sqrt{25-7t^2}}~dt

    2. \int\frac{x^3}{x^2-2x+3}~dx

    3. \int e^{3x}cos{4x}~dx
    1. Let \sqrt{7}t=5sin(\theta).
    3. By parts, u=cos(4x) and  dv= e^{3x}dx.
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  4. #4
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    Thanks for all your help!

    For #3, I've tried doing that, but I keep going around in circles:

    uv - \int vdu

    \frac{1}{3}cos(4x)e^{3x} - \frac{-4}{3}\int e^{3x}sin(4x)~dx

    What should I do?
    Last edited by greenstupor; February 7th 2010 at 07:24 PM.
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  5. #5
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    Oh, and I'm also not sure what you mean by your tip on #1. Sorry. I'm trying my best on this stuff.
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  6. #6
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    Quote Originally Posted by greenstupor View Post
    Thanks for all your help!

    For #3, I've tried doing that, but I keep going around in circles:

    uv - \int vdu

    \frac{1}{3}cos(4x)e^{3x} - \frac{-4}{3}\int e^{3x}sin(4x)~dx

    What should I do?
    Do by parts again.
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  7. #7
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    Quote Originally Posted by greenstupor View Post
    Oh, and I'm also not sure what you mean by your tip on #1. Sorry. I'm trying my best on this stuff.
    Did you apply this substitution on the integral?
    Do you the trigonometric substitutions?
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  8. #8
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    Quote Originally Posted by greenstupor View Post
    Thanks for all your help!

    For #3, I've tried doing that, but I keep going around in circles:

    uv - \int vdu

    \frac{1}{3}cos(4x)e^{3x} - \frac{-4}{3}\int e^{3x}sin(4x)~dx

    What should I do?
    If you integrate by parts again, you will get something like
    \int e^{3x}sin(4x)dx= A+ B\int e^{3x}sin(4x) dx.

    Solve that equation for \int e^{3x}sin(4x) dx
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