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March 24, 2017

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find the integral:

∫tan^6 (x) dx

  • calculus - ,

    Put tan(x) = u ---->

    x = arctan(u) ---->

    dx = du/(1+u^2)

    Integral of u^6/(1+u^2) du =

    Integral of

    (u^6 + u^4 - u^4 - u^2 + u^2 + 1 -1)/
    (u^2 + 1) du =

    Integral of

    (u^4 - u^2 + 1) du - Integral of
    1/(1+u^2) du =

    1/5 u^5 - 1/3 u^3 + u - arctan(u) + c=

    1/5 tan^5(x) - 1/3 tan^3(x) + tan(x) - x + c

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