Calculus

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Can someone help me to evaluate these two integrals?

INT dx/(x^2*sqrt(x^2-25))

and

INT (from 0 to 1) -5*sqrt(x^2+1) dx

  • Calculus -

    Run, do not walk to the section on trig substitutions. Usually that's the best tool for problems like these.

    Recall that sec^2 = 1 + tan^2, so that
    sec^2 - 1 = tan^2. Let x=5sec(u), so x^2-25 = 25sec^2(u)-25 = 25 tan^2(u)

    x^2 = 25 sec^2(u)
    dx = 5 sec(u) tan(u)

    Our integral now becomes

    5sec(u)tan(u)/(25sec^2(u) * 5tan(u)) = 1/25sec(u) du = cos(u)/25 du

    Int 1/25 cos(u) = 1/25 sin(u)

    Since x = 5 sec(u),
    x/5 = sec(u), and 5/x = cos(u)

    sin(u) = sqrt(1-25/x^2) = 1/x * sqrt(x^2-25), and our answer is thus

    1/25 sin(u) = sqrt(x^2-25)/25x

    *whew*

    The second one is similar, but uses hyperbolic functions:

    Let x = tan(u), so x^2+1 = sec^2(u)
    dx = sec^2(u) du

    and the integral becomes

    -5 Int sec^3(u) du

    Use integration by parts (twice) to get

    -5/2 tan(u)sec(u) - 5/2 log(tan u + sec u)
    = -5/2 x sqrt(x^2+1) - 5/2 log(x + sqrt(x^2+1)
    = -5/2[x sqrt(x^2+1) + arcsinh(x)]

    from 0 to 1
    At 0, the function is 0, so we just have f(1) = -5/2(sqrt(2) + asinh(1))
    = -5/2(1.414 + 2.435) = 4.62

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