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Thread: Integration of Rational Functions by partial fractions

  1. #16
    Eater of Worlds
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    With trig sub it whittlles down nicely.

    Let $\displaystyle x=sec({\theta}), \;\ dx=sec({\theta})tan({\theta})d{\theta}$

    Make the subs and you get:

    $\displaystyle \int_{0}^{\frac{\pi}{3}}tan^{2}({\theta})d{\theta}$
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  2. #17
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by galactus View Post
    With trig sub it whittlles down nicely.

    Let $\displaystyle x=sec({\theta}), \;\ dx=sec({\theta})tan({\theta})d{\theta}$

    Make the subs and you get:

    $\displaystyle \int_{0}^{\frac{\pi}{3}}tan^{2}({\theta})d{\theta}$
    yeah I did it that way, but I got stuck on the limits of integration I don't know what $\displaystyle 1 = sec({\theta}) $ is. Is there a quick way to figure out what they are?
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  3. #18
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    yeah I did it that way, but I got stuck on the limits of integration I don't know what $\displaystyle 1 = sec({\theta}) $ is. Is there a quick way to figure out what they are?
    $\displaystyle sec(\theta) = 1$

    $\displaystyle \frac{1}{cos(\theta)} = 1$

    $\displaystyle cos(\theta) = 1$

    $\displaystyle \theta = 0$

    -Dan
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  4. #19
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    Quote Originally Posted by FalconPUNCH! View Post
    I get $\displaystyle \frac{40}{3}$

    hmmm...seem to be getting the limits of integration wrong which is happening in my homework a lot.

    for this one $\displaystyle \int_{1}^{2} \frac{\sqrt{x^{2} - 1}}{x}dx $ is it best to use u substitution from the start?
    Alternatively, make the substitution $\displaystyle u^2 = x^2 - 1$.

    Then the integral becomes

    $\displaystyle I = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du$.

    Details can be given if needed .....
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  5. #20
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    Quote Originally Posted by mr fantastic View Post
    Alternatively, make the substitution $\displaystyle u^2 = x^2 - 1$.

    Then the integral becomes

    $\displaystyle I = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du$.

    Details can be given if needed .....
    $\displaystyle u^2 = x^2 - 1 \Rightarrow 2u \frac{du}{dx} = 2x \Rightarrow dx = \frac{u}{x} \, du$

    $\displaystyle x = 1 \Rightarrow u^2 = 0 \Rightarrow u = 0$. $\displaystyle x = 2 \Rightarrow u^2 = 3 \Rightarrow u = \sqrt{3}$ where the positive root is used.

    Then:

    $\displaystyle I = \int_{0}^{\sqrt{3}} \frac{u}{x} \frac{u}{x} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{x^2} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du$


    $\displaystyle = \int_{0}^{\sqrt{3}} \frac{(u^2 + 1) - 1}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du$,


    which integrates easily to give $\displaystyle I = \sqrt{3} - \frac{\pi}{3}$.
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  6. #21
    Member FalconPUNCH!'s Avatar
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    Thank you guys for helping me out. I'm going to use this thread as a reference so I won't forget how to do these.
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  7. #22
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    Quote Originally Posted by FalconPUNCH! View Post
    Thank you guys for helping me out. I'm going to use this thread as a reference so I won't forget how to do these.
    Good thinking, 99
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