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find the integral of x^2 - sec3xtan3x

  • Calculus -

    well spit it into 2 integrals
    integral x^2 dx = x^3/3

    sec^3 x tan x^3 dx
    = (1/cos^3)(sin^3/cos^3) dx
    =sin^3/cos^6 dx
    I can give you a recursion formula for this
    int dx sin ^m x/cos^n = sin^(m+1) x/[(n-1)cos^(n-1)x] - [(m-n+2)/(n-1)] int dx sin^m x/ cos^(n-2) x
    do that until you get to sin^3 x
    now int dx sin^3 x is -cos x + (1/3) cos^3 x

  • Calculus -

    or if you really meant

    the derivative of secx
    = (secx)tanx

    so the integral of sec3xtan3x = (1/3)sec3x

    so the integral of

    x^2 - sec3xtan3x
    (1/3)x^3 - (1/3)sec3x + C

    You have to be careful how you type these, as you can see they can be interpreted in different ways unless you use brackets to clearly state what you want.

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