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calculus

find the indefinite integral:
integration of tan^3(7x) dx

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john

  1. Yo can find a reduction formula for the integral of tan^n(x) as follows. We have that

    tan^n(x) = sin^n(x)/cos^n(x) =

    sin^(n-2)(x)/cos^n(x) sin^2(x) =

    sin^(n-2)(x)/cos^n(x) [1-cos^2(x)] =

    sin^(n-2)(x)/cos^n(x) -
    sin^(n-2)(x)/cos^(n-2)(x) =

    sin^(n-2)(x)/cos^(n-2)(x) 1/cos^2(x) -
    sin^(n-2)(x)/cos^(n-2)(x) =

    tan^(n-2)(x) 1/cos^2(x) - tan^(n-2)(x)

    Now, 1/cos^2(x) is the derivate of
    tan(x), so you can immediately integrate the first term:

    Integral of tan^(n-2)(x) 1/cos^2(x)dx =

    1/(n-1) tan^(n-1)(x)

    The integral of the second term is, of course, a similar problem as the original problem, but with a lower value for n, so you can interate the formula until you end up at n = 1 or n = 0.

    So, denoting the integral of tan^n(x)dx by I_n, we have:

    I_n = 1/(n-1) tan^(n-1)(x) - I_{n-2}

    For n = 3, we get:

    I_3 = 1/2 tan^2(x) - I_1

    and I_1 is the integral of tan(x)dx, which is -Log|cos(x)|.

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    Count Iblis

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