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Math Help - Trigonometric Integral

  1. #1
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    Trigonometric Integral

    How is tan^5 x integrated to get (1/4)sec^4(x) - tan^2(x) + ln|sec x| + C ?

    I tried integrating 2 different ways and was unable to get the books answer.
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Quote Originally Posted by Mike9182 View Post
    How is tan^5 x integrated to get (1/4)sec^4(x) - tan^2(x) + ln|sec x| + C ?

    I tried integrating 2 different ways and was unable to get the books answer.
    When integrating trigonometric integrals with odd powers, like tan^5 x, it can be useful to factor out a tan x and then use a trig identity dealing with even powers, specifically tan^2 x
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  3. #3
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    \int{\tan^5{x}\,dx} = \int{\tan^2{x}\tan^3{x}\,dx}

     = \int{(\sec^2{x} - 1)\tan^3{x}\,dx}

     = \int{\sec^2{x}\tan^3{x} - \tan^3{x}\,dx}

     = \int{\sec^2{x}\tan^3{x} - \tan^2{x}\tan{x}\,dx}

     = \int{\sec^2{x}\tan^3{x} - (\sec^2{x} - 1)\tan{x}\,dx}

     = \int{\sec^2{x}\tan^3{x} - \sec^2{x}\tan{x} + \tan{x}\,dx}

     = \int{\sec^2{x}(\tan^3{x} - \tan{x})\,dx} + \int{\tan{x}\,dx}

    = \int{\sec^2{x}(\tan^3{x} - \tan{x})\,dx} - \int{\frac{-\sin{x}}{\phantom{-}\cos{x}}\,dx}.


    Now integrate the first with a substitution u = \tan{x} and the second with a substitution u = \cos{x}.
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