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Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

How would i go about solving with trig sub?

Hi, ive been stumped by this integral as it does not contain a root, can anyone help?

=∫(-x^2+9) dx

thanks.

8 Answers

Relevance
  • 1 month ago

    ∫(-x^2+9) dx

    =- 1/3x^3 + 9x + C answer//

  • ?
    Lv 7
    1 month ago

    I=S(9-x^2)dx

    Let x=3sinA, then dx=3cosAdA

    I=9S[1-sin^2(A)](3cosAdA)

    =>

    I=27[S cosAdA-S sin^2(A)d(sinA)]

    =>

    I=27[sinA-sin^3(A)/3]+C

    =>

    I=27sinA[1-sin^2(A)/3]+C

    =>

    I=27sinAcos^2(A)/3+C

    If you want the result in terms of x, you

    should integrate directly with respect

    to x getting I=9x-(x^3)/3+C.

  • 1 month ago

    Hi, I'm sure you know how to solve it by the easiest method, although it is likely that you were asked to solve it with trig sub.

    So to solve it by trig sub in this case you can use a mathematical artifice. you artificially add a root and thus use trig sub normally.

    Attachment image
  • alex
    Lv 7
    1 month ago

    ∫ax^n dx = (a/(n+1))x^(n+1)+C , where n≠ -1

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  • ?
    Lv 7
    1 month ago

    So easy it's difficult!

    Integral of -x^2 dx is -x^3/3,

    integral of 9 dx is 9x,

    answer to your problem is 9x - (1/3)x^3 + C.

    Don't do "trig sub" !!!

  • ?
    Lv 7
    1 month ago

    You would directly integrate this. You would not use trig sub to do this. Although you can, you will just really really overcomplicate matters.

  • 1 month ago

    = - ∫(x^2-9) dx

    = -∫x^2dx + ∫9 dx

    = -x^3/3 + 9x +C

    stop looking for complications!

  • 1 month ago

    integral(-x^2 + 9) dx = 9 x - x^3/3 + constant

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