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Math Help - Need some help on a couple integrations

  1. #1
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    Need some help on a couple integrations

    You dont need to do them all but one would be nice. I want to see how to do it rather than just get answers. Thanks.

    Integral of [(e^x + cos(x)]^2 dx
    I tried to expand it and then got confused with integration by parts but I think you use that

    Integral of e^2x ln (1+e^x) dx
    I always hated ln

    Integral of (12-4x-x^2)^.5 dx
    I tried to factor it and got it to [-(x+6)(x-2)]^.5

    Integral of cos^4 (x) sin^3 (x) dx
    Should i change the cos^4 (x) to like (1-sin(x))^2 or something?

    Integral of [(4-x^2)^.5]/ x dx
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  2. #2
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    Quote Originally Posted by DaAzNJRiCh View Post

    Integral of e^2x ln (1+e^x) dx
    I always hated ln
    I assume,
    \int e^{2x}\ln (1+e^x)dx
    You can write it as,
    \int [(1+e^x)-1] \ln (1+e^x) e^x dx
    Then we can express this as,
    (f\circ g) \cdot g'
    Where,
    g=1+e^x
    f=(x-1)\ln x
    Thus we need to find,
    F=\int (x-1)\ln x dx
    Which you can do with integration by parts.
    Then by the subsitution rule the answer to your question is,
    F\circ g+C
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  3. #3
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    Quote Originally Posted by DaAzNJRiCh View Post
    You dont need to do them all but one would be nice. I want to see how to do it rather than just get answers. Thanks.

    Integral of [(e^x + cos(x)]^2 dx
    I tried to expand it and then got confused with integration by parts but I think you use that
    \int [e^x+\cos(x)]^2\ dx=\int e^{2x}+2e^x \cos(x)+(\cos(x))^2\ dx= <br />
\int e^{2x}\ dx +2\int e^x \cos(x)\ dx+\int (\cos(x))^2\ dx

    The first and last integrals on the right are standard integrals and should pose
    no problem. Now consider the middle integral on the right:

    <br />
2\int e^x \cos(x)\ dx=2\ Re \left[ \int e^x e^{ix}\ dx \right]=2\ Re \left[ \int e^{x(1+i)}\ dx \right]<br />

    Which is fairly simple to deal with.

    RonL
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  4. #4
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    Quote Originally Posted by DaAzNJRiCh View Post

    Integral of (12-4x-x^2)^.5 dx
    I tried to factor it and got it to [-(x+6)(x-2)]^.5
    Okay you have,
    \int \sqrt{12-4x-x^2} dx
    You factored, but you should have completed the square.
    \int \sqrt{-(x^2+4x-12)}dx
    Thus,
    \int \sqrt{-(x^2+4x+4-12-4)}dx
    Thus,
    \int \sqrt{16-(x+2)^2}dx
    Use a linear subsitution,
    u=x+2
    Thus,
    \int \sqrt{4^2-u^2} du
    This is a standard inverse sine function substitution.
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  5. #5
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    Quote Originally Posted by DaAzNJRiCh View Post

    Integral of cos^4 (x) sin^3 (x) dx
    Should i change the cos^4 (x) to like (1-sin(x))^2 or something?
    Almost!
    \int \cos^4 x\sin^2 x\sin x dx
    \int \cos^4 x(1-\cos^2 x)\sin xdx
    Let,
    u=\sin x
    Thus,
    \int u^4(1-u^2)du
    The rest is trivial.
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