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Thread: Integration (2)

  1. #1
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    Integration (2)

    Using the substitution x = 1/y or otherwise, find ∫ {1/[x (x^2 - 1)^0.5]} dx.

    Thank you!!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by Tangera View Post
    Using the substitution x = 1/y or otherwise, find ∫ {1/[x (x^2 - 1)^0.5]} dx.
    Substituting $\displaystyle x=1/y$ you should get $\displaystyle -\int \frac{\mathrm{d}y}{\sqrt{1-y^2}}$. Now recall that the derivative of $\displaystyle y\mapsto \arccos y $ is $\displaystyle y\mapsto \frac{-1}{\sqrt{1-y^2}}$ and you are done.
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  3. #3
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    $\displaystyle \int \frac{1}{x\sqrt{x^{2}-1}} \: dx$

    $\displaystyle x = \frac{1}{y} \: \: \Rightarrow \: \: dx = -\frac{1}{y^2} \: dy$

    Making the sub:
    $\displaystyle \int \frac{1}{\frac{1}{y}\sqrt{\left(\frac{1}{y}\right) ^{2} - 1}} \: \left(-\frac{1}{y^2} \: dy\right) \: \: = \: \: -\int \frac{1}{y\sqrt{\frac{1}{y^2} - 1}} \: dy$

    Now, imagine: $\displaystyle y = \sqrt{y^2}$

    $\displaystyle -\int \frac{1}{\sqrt{y^{2}}\sqrt{\frac{1}{y^2} - 1}} \: dy \: \: = \: \: -\int \frac{1}{\sqrt{1 - y^{2}}} \: dy$

    Does arccos/arcsin ring a bell?
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  4. #4
    Super Member wingless's Avatar
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    $\displaystyle \arccos \frac{1}{x}$ only works for positive x. Remember, $\displaystyle \sqrt{x^2} = |x| \color{red}\neq x$

    I have a more general solution.

    $\displaystyle \int \frac{1}{x\sqrt{1-x^2}}~dx$

    Now I'll do something that you'll understand the reason later.

    $\displaystyle \int \frac{x}{x^2\sqrt{1-x^2}}~dx$

    Let $\displaystyle u=\sqrt{x^2-1}$. Then $\displaystyle u^2 = x^2-1$. And $\displaystyle u~du = x~dx$. Now substitute everything,

    $\displaystyle \int \frac{u}{(u^2+1)u}~du = \int \frac{1}{u^2+1}~du = \arctan u+C$

    $\displaystyle = \arctan \sqrt{x^2-1}+C$
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