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Math Help - trigonometric substitution

  1. #1
    Member
    Joined
    Feb 2009
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    91

    trigonometric substitution

    Hi

    the integral is Integral sign ("S") dx/ (sqrt(21-4x-x^2))

    So first I completed the square so 21 - (x+2)^2 +4 so
    Let u = x+2 so du = dx

    so the original integral becomes S du/ sqrt( 25 - u^2)

    Then I let u = 5 sin t which makes sqrt (25 - u^2) = 5cos t
    du = 5 cos t dt

    so then S du/ sqrt(25 - u^2) = S 5 cos t dt / 5 cos t

    is this right???
    = S dt = t + C
    = sin ^-1 (u/5) +C = sin^-1 ((x+2)/5) + C

    I find the 5 sin t = u where t = sin^-1 u/5 so if this is wrong let me know.

    Thanks
    Calculus beginner
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  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    We may factor the denominator of the integrand:

    \frac{1}{21-4x-x^2}=\frac{1}{(x+7)(3-x)}.

    Now the method of Partial Fractions may be used to split the integrand into the sum of two fractions:

    \frac{1}{(x+7)(3-x)}=\frac{A}{x+7}+\frac{B}{3-x}.
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