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Math Help - integration (beginner)

  1. #1
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    integration (beginner)

    I've just started to learn about integration and can't even solve these examples:

    a) \int (x^2-\frac{1}{2\sqrt{x}})^2\, dx

    b) \int (sin\frac{x}{2}+cos\frac{x}{2})^2\, dx

    Could someone show me how to do it?
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  2. #2
    Senior Member Spec's Avatar
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    a) \int \left(x^2-\frac{1}{2\sqrt{x}}\right)^2\, dx=\int \left(x^4-x^{3/2}+\frac{1}{4x}\right)dx

    b) \int (sin\frac{x}{2}+cos\frac{x}{2})^2\, dx=\int \left(\sin^2 \frac{x}{2} +2\cos \frac{x}{2}\sin \frac{x}{2} +\cos^2 \frac{x}{2}\right)dx

    a) Should be easy to solve from the last step.

    b) Use the trigonometric identitities: 2\cos^2 \frac{x}{2} = 1+\cos x, 2\sin^2 \frac{x}{2} = 1-\cos x and note that 2\cos \frac{x}{2}\sin \frac{x}{2}=\frac{d}{dx}\left(2\sin^2\frac{x}{2} \right)
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  3. #3
    Super Member Deadstar's Avatar
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    I would maybe guess that you're having trouble with figuring out how to integrate them rather than expanding it so, continuing from Specs post.

    You can separate the integral in a) to be...
    \int x^4 dx - \int x^{3/2} dx + \int \frac{1}{4x} dx

    And then using the fact that \int x^n dx = \frac{x^{n+1}}{n+1}, you should be able to solve these.

    For part b) use \int \sin(x) dx = -\cos(x) and \int \cos(x) dx = \sin(x) after you've used the identities given in Specs post.
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  4. #4
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    Thank you!
    So a) \int \left(x^2-\frac{1}{2\sqrt{x}}\right)^2\, dx=\int \left(x^4-x^{3/2}+\frac{1}{4x}\right)dx=\frac{x^5}{5}-\frac{2\sqrt{x^5}}{5}+\frac{1}{4}ln|x|+C =\frac{x^5}{5}-\frac{2x^2\sqrt{x}}{5}+\frac{1}{4}ln|x|+C
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  5. #5
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    And for b) I got: x-cosx+C
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  6. #6
    Senior Member Spec's Avatar
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    EDIT: Nevermind, didn't notice that you rewrote the answer using the identity above.
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  7. #7
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    I've another 2 examples and I'm not sure how to work out them:

    c)  \int cos(\frac{3}{7}x+2)\, dx

    d)  \int \frac{2}{3x+5}\, dx

    For c) I tryed:

    \int cos(\frac{3}{7}x+2)\, dx =*

    \frac{3}{7}x+2=t

    d(\frac{3}{7}x+2)=dt

     \frac{3}{7}dx=dt

    dx=\frac{7dt}{3}

     *=\int cos \ t\frac{7dt}{3}=\frac{7}{3}\int cos \ t \, dt =\frac {7}{3} sin \ t+C=\frac{7}{3}sin(\frac{3}{7}x+2)+C

    Does it look correct?
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  8. #8
    Senior Member Spec's Avatar
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    c) looks correct.

    For d), use the fact that \frac{2}{3} \cdot \frac{d}{dx}\left( \ln(3x+5)\right)=\frac{2}{3} \cdot \frac{3}{3x+5}=\frac{2}{3x+5}
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  9. #9
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    One last question in this thread:

    e)  \int e^{2-0.2x}\,dx

    Is the answer  -5e^{2-0.2x}+C ?
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  10. #10
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    e^{2 - 0.2x} = e^2e^{-0.2x}

    Where e^2 is just a constant. Use a u substitution.

    e^2 \int e^{-0.2x}dx = \frac{-e^2}{0.2}e^{-0.2x} = \frac{-e^{2-0.2x}}{0.2}
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