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Math Help - Trig intergral help

  1. #1
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    Trig intergral help

    integrate: \frac{(sec2x)^2}{(tan2x)^5} dx


    After i convert to tangets and such and do the u substitution i end up with this

    u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3}) " alt="\frac{1}{2} integrate u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3}) " />

    I must have gone wrong somewhere becuase if you integrate 6u^{-1} you get 6\frac{u^{0}}{0}
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  2. #2
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    Quote Originally Posted by x5pyd3rx View Post
    integrate: \frac{(sec2x)^2}{(tan2x)^5} dx


    After i convert to tangets and such and do the u substitution i end up with this

    u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3}) " alt="\frac{1}{2} integrate u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3}) " />

    I must have gone wrong somewhere becuase if you integrate 6u^{-1} you get 6\frac{u^{0}}{0}
    You don't use the power law on x^-1 because you'd get the issue you described. Integrating x^{-1} is a special case:

    \int x^{-1} = \ln |x| + C
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  3. #3
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    so with those changes does this seem correct? \frac{1}{-6u^3} + \frac{4}{-4u^2} + 3\ln{u} + 4\frac{u^5}{10} + \frac{u^4}{8} +c
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by x5pyd3rx View Post
    integrate: \frac{(sec2x)^2}{(tan2x)^5} dx


    After i convert to tangets and such and do the u substitution i end up with this

    u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3}) " alt="\frac{1}{2} integrate u ^{-4} + 4u^{-3} + 6u^{-1} + 4u^{4} + u ^{3}) " />

    I must have gone wrong somewhere becuase if you integrate 6u^{-1} you get 6\frac{u^{0}}{0}
    How about this:

    \frac{sec^{2}(2x)}{tan^{5}(2x)} = \frac{1}{cos^{2}(2x)} \times \frac{1}{tan^{3}(2x)} \times \frac{1}{tan^{2}(2x)}

    = \frac{1}{cos^{2}(2x)} \times \frac{1}{tan^{3}(2x)} \times \frac{cos^{2}(2x)}{sin^{2}(2x)}

    = cot^{3}(2x) \times csc^{2} 2x

    now integrate

    \int cot^{3}(2x) \times csc^{2} 2x dx using substution rule..
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  5. #5
    Newbie Homeomorphism's Avatar
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    Quote Originally Posted by x5pyd3rx View Post
    so with those changes does this seem correct? \frac{1}{-6u^3} + \frac{4}{-4u^2} + 3\ln{u} + 4\frac{u^5}{10} + \frac{u^4}{8} +c
    Not sure what you have done here, but, for \int\dfrac{\left(\sec{2x}\right)^2}{\left(\tan{2x}  \right)^5}\;{dx}, if you let u = \tan{2x}, then \dfrac{du}{dx} = 2\sec^2{2x} \Rightarrow  dx = \dfrac{du}{2\left(\sec{2x}\right)^2}; so we have \int\dfrac{\left(\sec{2x}\right)^2}{\left(\tan{2x}  \right)^5}\;{dx} = \int\dfrac{\left(\sec{2x}\right)^2}{2u^5\left(\sec  {2x}\right)^2}\;{du}, which should be easier to finish off.
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