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Math Help - Evaluate the integral : secx/sqrt(sin2x)

  1. #1
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    Evaluate the integral : secx/sqrt(sin2x)

    Hello
    again i got i headache ;p

    Evaluate the integral:
    secx/sqrt(sin2x)

    sin(2x) not sin^2(x)
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  2. #2
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    Quote Originally Posted by TWiX View Post
    Hello
    again i got i headache ;p

    Evaluate the integral:
    secx/sqrt(sin2x)

    sin(2x) not sin^2(x)
    Mathematica gives the answer as \sec(x) \sqrt{\sin(2x)}

    Since the derivative of this would involve more than one term, I'm thinking integration by parts a logical choice but let me think some more about it. What have you tried?
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  3. #3
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     \frac{1}{\cos(x) \sqrt{\sin(2x)}}


     = \frac{1 + \cos(2x) - \cos(2x) }{\cos(x) \sqrt{\sin(2x)}}

     = \frac{2\cos(x)}{\sqrt{\sin(2x)}} - \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}}


    its integral  =


     \int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \int \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}} ~dx


     = \int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \sqrt{\sin(2x)}\sec(x) + \int \sqrt{\sin(2x)} \sec(x) \tan(x) ~dx

     = \int \frac{ 2\cos^2(x) + 2\sin^2(x)}{ \cos(x) \sqrt{\sin(2x)} } ~dx - \sqrt{\sin(2x)}\sec(x) +C

    therefore

     I = 2I - \sqrt{\sin(2x)}\sec(x) + C ~ , ~   I = \sqrt{\sin(2x)}\sec(x) + C'



    It is just a trial ... but luckily it is a successful trial
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  4. #4
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    Quote Originally Posted by Jameson View Post
    Mathematica gives the answer as \sec(x) \sqrt{\sin(2x)}

    Since the derivative of this would involve more than one term, I'm thinking integration by parts a logical choice but let me think some more about it. What have you tried?
    i tried by parts with u=secx
    but it was hard
    i tried to simplify it using trigonometric identities but it does not work
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  5. #5
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    Quote Originally Posted by simplependulum View Post
     \frac{1}{\cos(x) \sqrt{\sin(2x)}}


     = \frac{1 + \cos(2x) - \cos(2x) }{\cos(x) \sqrt{\sin(2x)}}

     = \frac{2\cos(x)}{\sqrt{\sin(2x)}} - \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}}


    its integral  =


     \int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \int \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}} ~dx


     = \int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \sqrt{\sin(2x)}\sec(x) + \int \sqrt{\sin(2x)} \sec(x) \tan(x) ~dx

     = \int \frac{ 2\cos^2(x) + 2\sin^2(x)}{ \cos(x) \sqrt{\sin(2x)} } ~dx - \sqrt{\sin(2x)}\sec(x) +C

    therefore

     I = 2I - \sqrt{\sin(2x)}\sec(x) + C ~ , ~ I = \sqrt{\sin(2x)}\sec(x) + C'



    It is just a trial ... but luckily it is a successful trial
    (1+cos(2x))/cosx = cosx??
    i didnt follow it :S
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  6. #6
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    Very nice job, simplependulum!

    Quote Originally Posted by TWiX View Post
    (1+cos(2x))/cosx = cosx??
    i didnt follow it :S
    I don't quite follow where you are point to. Do you mean the line (1+cos(2x)-cos(2x)), which is the second line in simplependulum's post?
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  7. #7
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    Quote Originally Posted by Jameson View Post
    Very nice job, simplependulum!


    I am just wondering if there is someone who can think of

     1 = \tan(x) \sin(2x) + \cos(2x) ( the numberator obtained when we differentiate  \sec(x) \sqrt{\sin(2x)} )

    ... i am sure that this guy is a genius !
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  8. #8
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    Quote Originally Posted by Jameson View Post
    Very nice job, simplependulum!



    I don't quite follow where you are point to. Do you mean the line (1+cos(2x)-cos(2x)), which is the second line in simplependulum's post?
    no
    i know its = adding zero

    am talking about the first fraction in the third line
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  9. #9
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    Quote Originally Posted by TWiX View Post
    no
    i know its = adding zero

    am talking about the first fraction in the third line
    \frac{1+cos(2x)}{cos(x)}

    =\frac{1+2cos ^2(x)-1}{cos(x)}

    =2 \cos(x)

    Quote Originally Posted by simplependulum View Post
    I am just wondering if there is someone who can think of

     1 = \tan(x) \sin(2x) + \cos(2x) ( the numberator obtained when we differentiate  \sec(x) \sqrt{\sin(2x)} )

    ... i am sure that this guy is a genius !
    I'm sure you are genius !!
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  10. #10
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    Here is another method :


     \int \frac{\sec(x)}{\sqrt{\sin(2x)}}~dx

     =  \int \frac{\sec^2(x) \cos(x) }{\sqrt{\sin(2x)}}~dx

     = \int \sec^2(x) \frac{ \cos(x) }{ \sqrt{2} \sqrt{\sin(x)} \sqrt{\cos(x)}}~dx

     = \frac{1}{\sqrt{2}} \int \sec^2(x) \frac{ 1}{\sqrt{\tan(x)}}~dx

    by substituting  t = \tan(x)

    i obtain

     I = \frac{1}{\sqrt{2}}2 \sqrt{\tan(x)} + C

     I = \sqrt{2 \sin(x) \cos(x)} \sec(x) + C
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