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Math Help - ln(x),cos(x),sin(x),ex.

  1. #1
    Super Member dhiab's Avatar
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    ln(x),cos(x),sin(x),ex.

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  2. #2
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    \int e^x\sqrt{e^x-1} dx

    make u = e^x-1 \Rightarrow \frac{du}{dx} = e^x

    giving \int \frac{du}{dx}\sqrt{u} dx

    = \int \sqrt{u} du

    = \int u^{\frac{1}{2}} du

    = \frac{2}{3}u^{\frac{3}{2}}+c

    = \frac{2}{3}(e^x-1)^{\frac{3}{2}} +c

    Now sub in terminals and subtract.
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  3. #3
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    All of these seem to be solvable using substitution.

    1) Try using u=\tan\left(\frac{x}{2}\right)

    2) u=e^x-1

    3) u=3x^2+2x-1
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  4. #4
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    Hello, dhiab!

    The third one is quite messy . . .


    \int^3_1 x\ln(3x^2+2x-1)\,dx

    By parts: . \begin{array}{ccccccc}u &=& \ln(3x^2+2x-1) & & dv&=& x\,dx \\ \\[-3mm] du &=& \dfrac{(6x+2)\,dx}{3x^2+2x-1} & & v &=& \frac{1}{2}x^2 \end{array}

    We have: . \tfrac{1}{2}x^2\ln(3x^2+2x-1) - \int\tfrac{1}{2}x^2\,\frac{6x+2}{3x^2+2x-1}\,dx

    . . . . . . =\;\tfrac{1}{2}x^2\ln(3x^2+2x-1) - \int\frac{3x^3 + x^2}{3x^2 + 2x - 1}\,dx

    . . . . . . = \;\tfrac{1}{2}x^2\ln(3x^2+2x-1) - \int\left[x -\tfrac{1}{3} + \frac{\frac{5}{3}x - \frac{1}{3}}{3x^2+2x-1}\right]\,dx .
    (long division)

    . . . . . . = \;\tfrac{1}{2}x^2\ln(3x^2+2x-1) - \int\left(x - \tfrac{1}{3}\right)\,dx - \tfrac{1}{3}\int\underbrace{\left[\frac{5x-1}{(x+1)(3x-1)}\right]}_{\text{partial fractions}}\,dx<br />

    Good luck!

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  5. #5
    Super Member dhiab's Avatar
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    thank you for all
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  6. #6
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    Quote Originally Posted by putnam120 View Post
    All of these seem to be solvable using substitution.

    1) Try using u=\tan\left(\frac{x}{2}\right)

    2) u=e^x-1

    3) u=3x^2+2x-1
    The third one can not be solved using a u-substitution because \frac{du}{dx} = 6x + 2, NOT x.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    The third one can not be solved using a u-substitution because \frac{du}{dx} = 6x + 2, NOT x.
    Yes I am aware of that, but you can note that \int_1^3 x\ln(3x^2+2x-1)dx=\frac{1}{6}\int_1^3 6x\ln(3x^2+2x-1)dx. Now you should be able to solve 2\int_1^3\ln(3x^2+2x-1)dx with integration by parts.

    Putting this all together it would be "easier" instead solve \int_1^3(6x+2)\ln(3x^2+2x-1)dx and then do the necessary algebraic manipulations to find the value of \int_1^3 x\ln(33x^2+2x-1)dx.


    As for the hint to number (1) things will work out as follows:

    dx=\frac{2du}{1+u^2}

    \cos(x)=\frac{1-u^2}{1+u^2}

    \sin(x)=\frac{2u}{1+u^2}

    You can verify these by drawing the necessary right triangle.
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