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Thread: Intergration With Trigonometry

  1. #1
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    Intergration With Trigonometry

    $\displaystyle \int \frac{\sec^5 \theta}{\tan^2\theta}d\theta$

    I'm having troubles starting this. I've tried substituting for tangent,splitting up the secants, and a couple other methods but I get stuck after a couple steps. How should i start out?
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  2. #2
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    Quote Originally Posted by twoteenine View Post
    $\displaystyle \int \frac{\sec^5 \theta}{\tan^2\theta}d\theta$

    I'm having troubles starting this. I've tried substituting for tangent,splitting up the secants, and a couple other methods but I get stuck after a couple steps. How should i start out?
    integral[((Sec[x])&# x5e;5)/((Tan[x]&#x29 ;^2)] - Wolfram|Alpha

    Click Show Steps.
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  3. #3
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    $\displaystyle \frac{ \sec^5(x) }{ \tan^2(x) } $

    $\displaystyle = \frac{ \sec^4(x) }{ \tan^2(x) } \cdot \sec(x) $

    $\displaystyle = \frac{ ( \tan^2(x) + 1)^2 }{ \tan^2(x) }\cdot \sec(x) $

    $\displaystyle = \frac{ \tan^4(x) + 2\tan^2(x) + 1}{\tan^2(x) } \cdot \sec(x)$

    $\displaystyle = [ \tan^2(x) + 2 + \frac{1}{\tan^2(x)} ] \cdot \sec(x) $

    $\displaystyle = \tan^2(x)\sec(x) + 2\sec(x) + \frac{1}{ \tan(x) \cdot \tan(x) \cdot \cos(x) } $

    $\displaystyle = \tan^2(x)\sec(x) + 2\sec(x) + \csc(x)\cot(x) $

    its integral $\displaystyle = \int \tan^2(x)\sec(x) ~dx + 2\int \sec(x)~dx + \int \csc(x)\cot(x) ~dx $

    $\displaystyle = \int \tan^2(x)\sec(x) ~dx + 2\int \sec(x)~dx - \csc(x) + C $

    Consider $\displaystyle I = \int \tan^2(x)\sec(x) ~dx $

    $\displaystyle I = \int [ \sec^3(x) - \sec(x) ]~dx $ $\displaystyle (1) $

    but integration by parts gives

    $\displaystyle I = \sec(x) \cdot \tan(x) - \int \sec^3 (x)~dx $

    therefore ,

    $\displaystyle \int \sec^3(x) ~dx = \sec(x) \tan(x) - I $ $\displaystyle (2)$

    By substituting $\displaystyle (2)$ into $\displaystyle (1) $

    we have

    $\displaystyle I = \sec(x) \tan(x) - I - \int \sec(x)~dx ~~~ \implies $

    $\displaystyle I = \frac{1}{2} \sec(x) \tan(x) - \frac{1}{2} \int \sec(x)~dx $

    The integral




    $\displaystyle \int \frac{ \sec^5(x) }{ \tan^2(x) } ~dx = $

    $\displaystyle \frac{1}{2} \sec(x) \tan(x) + \frac{3}{2} \int \sec(x)~dx - \csc(x) + C $

    $\displaystyle = \frac{1}{2} \sec(x) \tan(x) + \frac{3}{2} \cosh^{-1}[\sec(x)] - \csc(x) + C $
    Last edited by simplependulum; Mar 17th 2010 at 11:45 PM.
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