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Thread: Integration problem

  1. #1
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    Integration problem

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

    My attempt

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


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


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

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

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

    How do I proceed from here? I think I have to change the limits of integration in terms of $\displaystyle \theta$ instead of $\displaystyle y$.
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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.
    $\displaystyle \int_{1}^{e}\frac{dy}{y\sqrt{1 + (lny)^{2}}}$

    My attempt

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


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


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

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

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

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

    I think you have substituted wrong.

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

    $\displaystyle ln y = tan\theta$

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


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

    Also, you have to change the limits...

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

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

    Then your integration would be, $\displaystyle \int_{0}^{\frac{\pi}{4}}{\frac{sec^{2}\theta~d\the ta}{\sqrt{1+tan^{2}\theta}}}$
    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 $\displaystyle y = cos^{2}\theta$ to find the limits, by substituting 1 and e into y to get the two new limits in terms of $\displaystyle \theta$
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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, $\displaystyle \theta=tan^{-1}(lny)$

    $\displaystyle ln y = tan\theta$

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


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

    Also, you have to change the limits...

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

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

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

    $\displaystyle y = cos^{2}\theta$?

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

    $\displaystyle y = cos^{2}\theta$?

    Therefore, $\displaystyle dy = -2cos\theta sin\theta d\theta$
    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 $\displaystyle y = cos^{2}\theta$. Please recheck how you have differentiated.
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