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Thread: Trig-sub Integration

  1. #1
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    Trig-sub Integration


    Does anyone know how to do the above integral? I don't know what I should sub into the equation
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  2. #2
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    if you want a trig. substitution, then put $\displaystyle x=5\sin t.$

    or put $\displaystyle x=\frac1t.$
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    if you want a trig. substitution, then put $\displaystyle x=5\sin t.$

    or put $\displaystyle x=\frac1t.$
    thanks, but could you explain briefly how one would know to choose those substitutions?
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  4. #4
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    Hello, cdlegendary!

    How can you not know?

    This is the very first trig substitution you ever learned . . .


    $\displaystyle \int\frac{-2\,dx}{x^2\sqrt{25-x^2}} $

    Let: .$\displaystyle x \:=\:5\sin\theta \quad\Rightarrow\quad dx \:=\:5\cos\theta\,d\theta$

    Also: .$\displaystyle \sqrt{25-x^2} \:=\:5\cos\theta$


    Substitute: .$\displaystyle -2\int\frac{5\cos\theta\,d\theta}{25\sin^2\!\theta\ cdot 5\cos\theta} \;=\;-\frac{2}{25}\int\frac{d\theta}{\sin^2\!\theta} \;=\;-\frac{2}{25}\int\csc^2\!\theta\,d\theta$


    Can you finish it now?

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  5. #5
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    let's make it more generally:

    suppose you have an expression in the integrand, say $\displaystyle \sqrt{a^2-x^2},$ so we want to delete the square root in order to make the integration easy, so in this case we invoke the trig. identity $\displaystyle \sin^2t+\cos^2t=1.$ Now if we put $\displaystyle x=a\sin t$ then $\displaystyle a^2-x^2=a^2-a^2\sin^2t=a^2(1-\sin^2t)=a^2\cos^2t,$ so yes, it's just taking the square root of a squared thing.
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  6. #6
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    Thanks for the help guys, its much appreciated. I can figure it out from here.
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