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Math Help - Trig Substitution

  1. #1
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    Trig Substitution

    The problem:
    integrate 1/sqr(16-x^2) dx


    My answer: (@ = theta) (trig substitution)
    x=sin @ a = 16 dx = cos @ sqr(16-x^2) = cos @
    = integrate cos @ / cos @ = 1

    The book's answer:
    arcsin(x/4)+c

    Where did I go wrong?
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  2. #2
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    Quote Originally Posted by Retromingent View Post
    The problem:
    integrate 1/sqr(16-x^2) dx


    My answer: (@ = theta) (trig substitution)
    x=sin @ a = 16 dx = cos @ sqr(16-x^2) = cos @
    = integrate cos @ / cos @ = 1

    The book's answer:
    arcsin(x/4)+c

    Where did I go wrong?
    \int \frac{1}{\sqrt{a^2 - x^2}}\, dx = \arcsin{\frac{x}{a}} + C.

    In this case,  a = 4 .

    So \int \frac{1}{\sqrt{16 - x^2}}\, dx = \int \frac{1}{\sqrt{4^2 - x^2}}\, dx = \arcsin {\frac{x}{4}} + C.
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