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Math Help - trig substitution

  1. #1
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    Exclamation trig substitution

    hey there

    I can't get this problem I've been trying it for a while now and it is really frustrating me...some help would be greatly appreciated

    integral of 5 / sqrt (48-16x-64x^2) dx

    I tried completing the square by getting 49-64(x+1/8)^2...is this right? I've seen so many ways to complete the square I don't know which I should use for integrals...

    and then I went ahead and tried to solve it by trig substitution but I just can't get it :S I used u= 7sint

    thanking you in advance
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  2. #2
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    Integral

    Hello zaddie21
    Quote Originally Posted by zaddie21 View Post
    hey there

    I can't get this problem I've been trying it for a while now and it is really frustrating me...some help would be greatly appreciated

    integral of 5 / sqrt (48-16x-64x^2) dx

    I tried completing the square by getting 49-64(x+1/8)^2...is this right? I've seen so many ways to complete the square I don't know which I should use for integrals...

    and then I went ahead and tried to solve it by trig substitution but I just can't get it :S I used u= 7sint

    thanking you in advance
    Have you noticed that you can take out a common factor of 16? It makes the numbers easier to handle.

    48-16x-64x^2 = 16(3-x-4x^2)

    ... and there's really only one way to complete the square, and that is:

    3 - x- 4x^2 = -4(x^2 + \tfrac{1}{4}x - \tfrac{3}{4})

    = -4[(x+\tfrac{1}{8})^2- \tfrac{3}{4}-\tfrac{1}{64}]

    = -4[(x+\tfrac{1}{8})^2-\tfrac{49}{64}]

    = 4[\tfrac{49}{64}-(x+\tfrac{1}{8})^2]

    So 48-16x-64x^2 = 16\cdot 4[\tfrac{49}{64}-(x+\tfrac{1}{8})^2]

    =64[(\tfrac{7}{8})^2 -(x+\tfrac{1}{8})^2]

    Put this into your integral, and you'll have something in the form

    \int\frac{kdx}{\sqrt{a^2 - (x+b)^2}}

    If you now substitute (x+b) = a\sin\theta, that should do it.

    Grandad
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