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Thread: Indefinite Integral Problem

  1. #1
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    Indefinite Integral Problem

    I would like some help with this indefinite integral problem.

    $\displaystyle \int\frac{6dx}{\sqrt{4-(x+1)^2}}$

    I know that the answer is going to have arcsin in it, but I don't know what to make u. I've tried $\displaystyle (x+1)$, but I don't know how to get rid of the 4. Any assistance on this problem would be greatly appreciated. Thank you.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by xwanderingpoetx View Post
    I would like some help with this indefinite integral problem.

    $\displaystyle \int\frac{6dx}{\sqrt{4-(x+1)^2}}$

    I know that the answer is going to have arcsin in it, but I don't know what to make u. I've tried $\displaystyle (x+1)$, but I don't know how to get rid of the 4. Any assistance on this problem would be greatly appreciated. Thank you.
    Note that $\displaystyle \int\frac{6\,dx}{\sqrt{4-(x+1)^2}}=\int\frac{6\,dx}{\sqrt{4\left(1-\left(\tfrac{x+1}{2}\right)^2\right)}}=\int\frac{3 \,dx}{\sqrt{1-\left(\tfrac{x+1}{2}\right)^2}}$

    Now apply the substitution $\displaystyle u=\frac{x+1}{2}$. Then you'll see the integral equation for arcsine.

    Does this make sense?
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  3. #3
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    Yes that makes sense. I didn't realize I could actually pull a 4 out of the $\displaystyle (x+1)^2$. Thank you very much.
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    Note that $\displaystyle \int\frac{6\,dx}{\sqrt{4-(x+1)^2}}=\int\frac{6\,dx}{\sqrt{4\left(1-\left(\tfrac{x+1}{2}\right)^2\right)}}=\int\frac{3 \,dx}{\sqrt{1-\left(\tfrac{x+1}{2}\right)^2}}$
    Actually, it was done, the only thing you needed to do here, was to put $\displaystyle x+1=2t$ and you have the arcsine.
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