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Math Help - do u wanna solve perfect ıntegration questions?:)

  1. #1
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    do u wanna solve perfect ıntegration questions?:)

    1) INT (arctane^x/2)/[e^x/2(1+e^x)]dx ??????


    2)INT dx/[x(1+x^3+x^6)^1/2]????????


    3)INT (e^2x)/(x^2-3x+2)dx ???????


    4)INT (ch^2)(sh^2)dx????????


    5)INT (ch^4)dx ????????


    please help meeee
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by berkanatci View Post
    1) INT (arctane^x/2)/[e^x/2(1+e^x)]dx ??????


    2)INT dx/[x(1+x^3+x^6)^1/2]????????


    3)INT (e^2x)/(x^2-3x+2)dx ???????


    4)INT (ch^2)(sh^2)dx????????


    5)INT (ch^4)dx ????????


    please help meeee
    Make sure you write these accurately:

    INT (arctane^x/2)/[e^x/2(1+e^x)]dx

    Do you mean:
    INT arctan(x/2)/[e^(x/2)*(1 + e^x)] dx ?

    Go through all the problems that you need help on and make sure they are written accurately, otherwise I can't help you if I'm solving the wrong problem.
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  3. #3
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    Quote Originally Posted by ecMathGeek View Post
    Make sure you write these accurately:

    INT (arctane^x/2)/[e^x/2(1+e^x)]dx

    Do you mean:
    INT arctan(x/2)/[e^(x/2)*(1 + e^x)] dx ?

    Go through all the problems that you need help on and make sure they are written accurately, otherwise I can't help you if I'm solving the wrong problem.

    the correct one is:INT [arctane^(x/2)]/[e^(x/2)*(1+e^x)]dx?

    I wrote it correctly
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by berkanatci View Post
    1) INT (arctane^x/2)/[e^x/2(1+e^x)]dx ??????
    Help is a little late, but here's number 1. And please use parenthesis!

    \int atn (e^{x/2} ) \frac{e^{x/2}}{1 + e^x} dx

    I'll transform the integral in steps:
    Let y = e^{x/2} \implies dy = \frac{e^{x/2}}{2} dx

    Thus
    \int atn (e^{x/2} ) \frac{e^{x/2}}{1 + e^x} dx = \int atn(y) \frac{2dy}{1 + y^2} = 2 \int atn(y) \frac{dy}{1 + y^2}

    Let z = atn(y) \implies dz = \frac{dy}{1 + y^2}

    Finally:
    \int atn (e^{x/2} ) \frac{e^{x/2}}{1 + e^x} dx = 2 \int atn(y) \frac{dy}{1 + y^2} = 2 \int z dz = z^2 + C

    = atn^2(y) + C =  atn^2 \left ( e^{x/2} \right ) + C

    -Dan
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  5. #5
    MHF Contributor red_dog's Avatar
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    2) \displaystyle\int\frac{dx}{x(1+x^3+x^6)^{\frac{1}{  2}}}=\int\frac{x^2dx}{x^3(1+x^3+x^6)^{\frac{1}{2}}  }
    Let x^3=t\Rightarrow 3x^2dx=dt
    Then I=\frac{1}{3}\int\frac{dt}{t\sqrt{t^2+t+1}}
    Now \displaystyle t=\frac{1}{u}\Rightarrow dt=-\frac{1}{u^2}du
    \displaystyle I=-\frac{1}{3}\int\frac{du}{\sqrt{u^2+u+1}}=-\frac{1}{3}\ln\left(u+\frac{1}{2}+\sqrt{u^2+u+1}\r  ight)+C
    Now back substitute.

    4) I=\int ch^2xsh^2xdx=\displaystyle\frac{1}{4}\int sh2xdx=\frac{1}{8}ch2x+C

    5) \int ch^4xdx=\int ch^2x(1+sh^2x)dx=\int ch^2xdx+\int ch^2xsh^2xdx=
    \displaystyle =\frac{1}{2}\int(1+ch2x)dx+\frac{1}{8}ch2x=\frac{x  }{2}+\frac{1}{4}sh2x+\frac{1}{8}ch2x+C
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by red_dog View Post
    2) \displaystyle\int\frac{dx}{x(1+x^3+x^6)^{\frac{1}{  2}}}=\int\frac{x^2dx}{x^3(1+x^3+x^6)^{\frac{1}{2}}  }
    Let x^3=t\Rightarrow 3x^2dx=dt
    Then I=\frac{1}{3}\int\frac{dt}{t\sqrt{t^2+t+1}}
    Now \displaystyle t=\frac{1}{u}\Rightarrow dt=-\frac{1}{u^2}du
    \displaystyle I=-\frac{1}{3}\int\frac{du}{\sqrt{u^2+u+1}}=-\frac{1}{3}\ln\left(u+\frac{1}{2}+\sqrt{u^2+u+1}\r  ight)+C
    Now back substitute.
    Oh that's just too cute!

    -Dan
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Oh that's just too cute!

    -Dan
    Completely agree. excellent work red_dog!
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  8. #8
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    Quote Originally Posted by berkanatci View Post
    5)INT (ch^4)dx ????????
    I assume it means,
    \int \cosh^4 x dx

    You can use the identity, \cosh^2 x = \frac{\cosh 2x + 1}{2} twice.

    4)INT (ch^2)(sh^2)dx????????
    \int \cosh^2 x \sinh^2 x dx
    Use "half angle identiy":
    \int \left( \frac{\cosh 2x + 1}{2} \right) \left( \frac{\cosh 2x - 1}{2} \right) dx

    The details are easy.
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  9. #9
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    Quote Originally Posted by berkanatci View Post
    3)INT (e^2x)/(x^2-3x+2)dx ???????
    Write,
    \int \frac{e^{2x}}{(x-2)} - \frac{e^{2x}}{(x-1)} dx

    Consider the first integral,
    \int \frac{e^{2x}}{x-2} dx let t = x-2 \implies t'=1.
    Thus,
    \int \frac{e^{2t}e^2}{t} dt = \int \frac{2e^{2t}e^2}{2t} dt = e^2\mbox{Ei}(2t)+C = e^2\mbox{Ei}(2x-4)+C

    Similarly do the other one.
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