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March 28, 2017

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evaluate the integral:

integral from -pi/4 to 0 for the function 6sec^3x dx.

it has to be an exact answer and i did it and keep getting it wrong. I got
4sqrt(2)-4ln(-sqrt(2)+1)

  • Calculus integral - ,

    According to the Wolfram integrator , this looks like a messy integration
    http://integrals.wolfram.com/index.jsp?expr=6%2F%28cos%28x%29%29%5E3&random=false

    looks like multiple integration by parts, arhhh!

  • Calculus integral - ,

    u = sec x,
    dv = sec^2 x dx

    du = secx tanx dx
    v = tan x

    ∫sec^3 x dx
    = ∫u dv = uv - ∫v du
    = secx tanx - ∫secx tan^2 x dx
    = secx tanx - ∫(secx (sec^2 x - 1) dx
    = secx tanx - ∫sec^3 x dx + ∫secx dx
    so,
    2∫sec^3 x dx = secx tanx + ∫secx dx
    = secx tanx + ln(secx tanx)

    ∫sec^3 x dx = 1/2 (secx tanx + ln(secx + tanx)

    plug in 0 and pi/4 to get

    1/2 (√2*1 + ln(√2+1)) - 1/2(1*0 + ln(1+0))
    = 1/2 (√2 + ln(√2+1))

    multiply by 6 to get 3(√2 + ln(√2+1))

    double check my math, and you will either
    (a) see your mistake
    (b) see my mistake

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