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Math Help - Integration problem?

  1. #1
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    Integration problem?

    How in the world do you solve this???:

    int[sec(1-x)*tan(1-x)*dx]

    Thank you!
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  2. #2
    Senior Member nikhil's Avatar
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    Red face Try it

    Hint

    int(sec(x)tan(x))dx=sec(x)+c
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  3. #3
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    Ok,

    so using u-substitution:

    u = sec(1-x)
    du = sec(1-x)*tan(1-x)?

    Thanks!
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  4. #4
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    Quote Originally Posted by dillonmhudson View Post
    How in the world do you solve this???:

    int[sec(1-x)*tan(1-x)*dx]

    Thank you!
    Remember that \sec{(1 - x)} = \frac{1}{\cos{(1 - x)}} and that \tan{(1 - x)} = \frac{\sin{(1-x)}}{\cos{(1 -x)}}.

    Using these, we see that

    \int{\sec{(1-x)}\,\tan{(1-x)}\,dx} = \int{\frac{1}{\cos^2{(1-x)}}\,\sin{(1-x)}\,dx}.

    Use the substitution u = \cos{(1-x)} so that \frac{du}{dx} = \sin{(1-x)}. Then the integral becomes

    \int{\frac{1}{u^2}\,\frac{du}{dx}\,dx}

     = \int{u^{-2}\,du}

     = -u^{-1} + C

     = -\frac{1}{\cos{(1-x)}} + C

     = -\sec{(1-x)} + C.
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  5. #5
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    Awesome!

    Thank you both very much!
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