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Thread: How to Integrate this?

  1. #1
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    How to Integrate this?

    sec2xdx/cos^2[2x](sex2x-1)


    Thanks!
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  2. #2
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    Quote Originally Posted by honestliar View Post
    sec2xdx/cos^2[2x](sex2x-1)


    Thanks!
    $\displaystyle Intg$( sec^3[2x])$\displaystyle div$(sec2x - 1)
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  3. #3
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    Hello, honestliar!

    sa-ri-ga-ma gave you a big hint?
    Was it enough?


    Integrate: .$\displaystyle \int \frac{\sec(2x)\,dx}{\cos^2(2x)[\sec(2x)-1]}$
    Since $\displaystyle \frac{1}{\cos(2x)} \:=\:\sec(2x)$

    . . we have: .$\displaystyle \int\frac{\sec^3(2x)\,dx}{\sec(2x)-1} \;=\;\int\bigg[\sec^2(2x) + \sec(2x) + 1 + \frac{1}{\sec(2x) - 1}\bigg]\,dx$

    The first three terms have standard integration formulas.


    For the fraction: .$\displaystyle \frac{1}{\sec(2x) - 1} \;=\;\frac{1}{\frac{1}{\cos(2x)} - 1} $

    Multiply by $\displaystyle \frac{\cos(2x)}{\cos(2x)}\!:\quad \frac{\cos(2x)}{\cos(2x)}\cdot\frac{1}{\frac{1}{\c os(2x)} - 1} \;=\;\frac{\cos(2x)}{1 - \cos(2x)}$

    Multiply by $\displaystyle \frac{1+\cos(2x)}{1+\cos(2x)}\!:\quad \frac{1+\cos(2x)}{1+\cos(2x)}\cdot \frac{\cos(2x)}{1-\cos(2x)} \;=\;\frac{\cos(2x) - \cos^2(2x)}{1 - \cos^2(2x)}$

    . . $\displaystyle =\;\;\frac{\cos(2x) - \cos^2(2x)}{\sin^2(2x)} \;\;=\;\;\frac{\cos(2x)}{\sin^2(2x)} - \frac{\cos^2(2x)}{\sin^2(2x)} \;\;=\;\;\frac{1}{\sin(2x)}\cdot\frac{\cos(2x)}{\s in(2x)} - \left(\frac{\cos(2x)}{\sin(2x)}\right)^2$

    . . $\displaystyle = \;\;\csc(2x)\cot(2x) - \cot^2(2x) \;=\;\csc(2x)\cot(2x) - [\csc^2(2x) - 1]$


    And we have: .$\displaystyle \int\bigg[\csc(2x)\cot(2x) - \csc^2(2x) + 1\bigg]\,dx$

    . . which also have standard integration formulas.

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  4. #4
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    Thank you very much, I got it
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  5. #5
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    Or

    $\displaystyle \int {\frac{{\sec 2x}}{{{{\cos }^2}2x\left( {\sec 2x - 1} \right)}}dx} .$

    $\displaystyle \frac{{\sec 2x}}{{{{\cos }^2}2x\left( {\sec 2x - 1} \right)}} = \frac{1}{{\cos 2x}}\frac{1}{{{{\cos }^2}2x\left( {\frac{{1 - \cos 2x}}{{\cos 2x}}} \right)}} = \frac{1}{{{{\cos }^2}2x\left( {1 - \cos 2x} \right)}} =$

    $\displaystyle = \frac{{1 - \cos 2x + \cos 2x}}{{{{\cos }^2}2x\left( {1 - \cos 2x} \right)}} = \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x\left( {1 - \cos 2x} \right)}} =$

    $\displaystyle = \frac{1}{{{{\cos }^2}2x}} + \frac{{1 - \cos 2x + \cos 2x}}{{\cos 2x\left( {1 - \cos 2x} \right)}} = \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x}} + \frac{1}{{1 - \cos 2x}} =$

    $\displaystyle = \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x}} + \frac{{1 + \cos 2x}}{{1 - {{\cos }^2}2x}} = \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x}} + \frac{{1 + \cos 2x}}{{{{\sin }^2}2x}} =$

    $\displaystyle = \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x}} + \frac{1}
    {{{{\sin }^2}2x}} + \frac{{\cos 2x}}{{{{\sin }^2}2x}}.$

    $\displaystyle \int {\frac{{dx}}{{{{\cos }^2}2x}}} = \frac{1}{2}\int {\frac{{d\left( {2x} \right)}}{{{{\cos }^2}2x}}} = \frac{1}{2}\tan 2x + C.$

    $\displaystyle \int {\frac{{dx}}{{\cos 2x}}} = \frac{1}{2}\int {\sec 2xd\left( {2x} \right)} = \frac{1}{2}\ln \left| {\tan 2x + \sec 2x} \right| + C.$

    $\displaystyle \int {\frac{{dx}}{{{{\sin }^2}2x}}} = \frac{1}{2}\int {\frac{{d\left( {2x} \right)}}{{{{\sin }^2}2x}}} = - \frac{1}{2}\cot 2x + C.$

    $\displaystyle \int {\frac{{\cos 2x}}{{{{\sin }^2}2x}}dx} = \left\{ \begin{gathered}\sin 2x = u, \hfill \\\cos 2xdx = \frac{{du}}{2} \hfill \\ \end{gathered} \right\} = \frac{1}{2}$$\displaystyle \int {\frac{{du}}{{{u^2}}}} = - \frac{1}{{2u}} + C = - \frac{1}{{2\sin 2x}} + C = - \frac{1}{2}\csc 2x + C.$

    Finally we have

    $\displaystyle \int {\frac{{\sec 2x}}{{{{\cos }^2}2x\left( {\sec 2x - 1} \right)}}dx} = \frac{1}{2}\tan 2x - \frac{1}{2}\cot 2x - \frac{1}{2}\csc 2x + \frac{1}{2}\ln$$\displaystyle \left| {\tan 2x + \sec 2x} \right| + C.$
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