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Math Help - Integration problem

  1. #1
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    Integration problem

    Use appropriate substitution and than a trigonometric substitution and evaluate the integral.
    \int_{1}^{e}\frac{dy}{y\sqrt{1 + (lny)^{2}}}

    My attempt

    \int_{1}^{e}\frac{dy}{y\sqrt{1 + (lny)^{2}}}


    ln y = tan\theta
    y = cos^{2}\theta
    dy = -2cos\theta sin\theta d\theta


    = -2\int_{1}^{e}\frac{cos\theta sin\theta d\theta}{cos^{2}\theta\sqrt{1 + tan^{2}\theta}}

    = -2\int_{1}^{e}\frac{sin\theta d\theta}{cos\theta sec\theta}

    = -2\int_{1}^{e}sin\theta d\theta

    How do I proceed from here? I think I have to change the limits of integration in terms of \theta instead of y.
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  2. #2
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    Substitute ln(y) = t.
    1/y*dy = dt.
    Substitute the limits of y and find the limit of t.
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  3. #3
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    Quote Originally Posted by temaire View Post
    Use appropriate substitution and than a trigonometric substitution and evaluate the integral.
    \int_{1}^{e}\frac{dy}{y\sqrt{1 + (lny)^{2}}}

    My attempt

    \int_{1}^{e}\frac{dy}{y\sqrt{1 + (lny)^{2}}}


    ln y = tan\theta
    y = cos^{2}\theta
    dy = -2cos\theta sin\theta d\theta


    = -2\int_{1}^{e}\frac{cos\theta sin\theta d\theta}{cos^{2}\theta\sqrt{1 + tan^{2}\theta}}

    = -2\int_{1}^{e}\frac{sin\theta d\theta}{cos\theta sec\theta}

    = -2\int_{1}^{e}sin\theta d\theta

    How do I proceed from here? I think I have to change the limits of integration in terms of \theta instead of y.
    Dear temaire,

    I think you have substituted wrong.

    Substitute, \theta=tan^{-1}(lny)

    ln y = tan\theta

    \frac{1}{y}\frac{dy}{d\theta}=sec^{2}\theta


    \frac{dy}{y}=sec^{2}{\theta}~d{\theta}

    Also, you have to change the limits...

    When, y=1\Rightarrow{\theta=0}

    When, y=e\Rightarrow{\theta=\frac{\pi}{4}}

    Then your integration would be, \int_{0}^{\frac{\pi}{4}}{\frac{sec^{2}\theta~d\the  ta}{\sqrt{1+tan^{2}\theta}}}
    Can you do it form here???
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  4. #4
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Substitute ln(y) = t.
    1/y*dy = dt.
    Substitute the limits of y and find the limit of t.
    Shouldn't I use y = cos^{2}\theta to find the limits, by substituting 1 and e into y to get the two new limits in terms of \theta
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    Dear temaire,

    I think you have substituted wrong.

    Substitute, \theta=tan^{-1}(lny)

    ln y = tan\theta

    \frac{1}{y}\frac{dy}{d\theta}=sec^{2}\theta


    \frac{dy}{y}=sec^{2}{\theta}~d{\theta}

    Also, you have to change the limits...

    When, y=1\Rightarrow{\theta=0}

    When, y=e\Rightarrow{\theta=\frac{\pi}{4}}

    Then your integration would be, \int_{0}^{\frac{\pi}{4}}{\frac{sec^{2}\theta~d\the  ta}{\sqrt{1+tan^{2}\theta}}}
    Can you do it form here???
    But if \frac{1}{y}\frac{dy}{d\theta}=sec^{2}\theta, isn't

    y = cos^{2}\theta?

    Therefore, dy = -2cos\theta sin\theta d\theta
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  6. #6
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    Quote Originally Posted by temaire View Post
    But if \frac{1}{y}\frac{dy}{d\theta}=sec^{2}\theta, isn't

    y = cos^{2}\theta?

    Therefore, dy = -2cos\theta sin\theta d\theta
    Dear temaire,

    You have to use the substitution you use to find the limits. I think you got it confused since your differentiation was wrong. You cannot obtain y = cos^{2}\theta. Please recheck how you have differentiated.
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