integral: sqrt (16 + 6x − x^2)

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asked by sam
  1. Note that since (x-3)^2 = x^2-6x+9, what you have is

    ∫√(25-(x-3)^2) dx
    If u = x-3, du = dx, and it's just
    ∫√(25-u^2) du

    which you should know. If not, just use the trig substitution u = 5sinθ

    The answer can be seen at^2%29+du

    Or, in terms of the original integral,^2%29+dx

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    posted by Steve
  2. thanks. i wish i could upload a screen shot of what i typed it but i did it exactly as the site gave and it was counted wrong.

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    posted by sam
  3. nevermind! i just forgot the constant

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    posted by sam

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