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Math Help - 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!
    Intg( sec^3[2x]) 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: . \int \frac{\sec(2x)\,dx}{\cos^2(2x)[\sec(2x)-1]}
    Since \frac{1}{\cos(2x)} \:=\:\sec(2x)

    . . we have: . \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: . \frac{1}{\sec(2x) - 1} \;=\;\frac{1}{\frac{1}{\cos(2x)} - 1}

    Multiply by \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 \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)}

    . . =\;\;\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

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


    And we have: . \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

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

    \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)}} =

    = \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)}} =

    = \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}} =

    = \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}} =

    = \frac{1}{{{{\cos }^2}2x}} + \frac{1}{{\cos 2x}} + \frac{1}<br />
{{{{\sin }^2}2x}} + \frac{{\cos 2x}}{{{{\sin }^2}2x}}.

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

    \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.

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

    \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} \int {\frac{{du}}{{{u^2}}}}  =  - \frac{1}{{2u}} + C =  - \frac{1}{{2\sin 2x}} + C =  - \frac{1}{2}\csc 2x + C.

    Finally we have

    \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 \left| {\tan 2x + \sec 2x} \right| + C.
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