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Math Help - Trig Substitution

  1. #1
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    Trig Substitution

    Integrate (√(1+x^2))/(x))

    I get to the integral of 1/(sinxcos²x)
    What would I do next
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  2. #2
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    use u=cosx and then do another trig substitution if you get stuck.
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  3. #3
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    Quote Originally Posted by andrewho93 View Post
    Integrate (√(1+x^2))/(x))
    Without a trigonometric substitution, doing in this way yields a very nice answer:

    \displaystyle u = \sqrt{1+x^2} \Leftrightarrow dx = \frac{\sqrt{1+x^2}}{x}\;{du} \displaystyle \Rightarrow \int\frac{\sqrt{1+x^2}}{x}\;{dx} = \int\left(\frac{\sqrt{1+x^2}}{x}\right)\left(\frac  {\sqrt{1+x^2}}{x}\right)\;{du} = \int\frac{u^2}{u^2-1}\;{du} [LaTeX ERROR: Convert failed] " alt=" \displaystyle = \int\frac{u^2}{(u+1)(u-1)}\;{du} = \int \left\{\frac{1}{2(u-1)}-\frac{1}{2(u+1)}+1 \right\}\;{du}[/Math] [LaTeX ERROR: Convert failed] " />[LaTeX ERROR: Convert failed]  [Math]\displaystyle = \boxed{\ln\sqrt{\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}}+\sqrt{1+x^2}+k}.

    I get to the integral of 1/(sinxcos²x)
    What would I do next
    Hint: [LaTeX ERROR: Convert failed] .
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  4. #4
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    oh that's a very smart hint
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  5. #5
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    i like when trig. is not everything, here's another way to tackle it:

    consider \displaystyle\int{\frac{dx}{x\sqrt{1+{{x}^{2}}}}}, substitute x=\dfrac1t and the integral becomes -\displaystyle\frac{dt}{\sqrt{1+t^2}}=-\ln\big(t+\sqrt{1+t^2}\big)+c_1, back substitute and get \displaystyle\int{\frac{dx}{x\sqrt{1+{{x}^{2}}}}}=  \ln \left( \frac{x}{1+\sqrt{1+{{x}^{2}}}} \right)+\sqrt{1+{{x}^{2}}}+k.
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