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Math Help - Integration problem

  1. #1
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    Integration problem

    Hello everyone,
    I've been preparing for my cal exam and I got stuck on this problem:
    Evaluate the infinite integral
    \int \frac {\sec^4 x}{\sqrt(tan 2x)} dx
    I would really appreciate if you give me a clue on how I should do it because I've been trying different trigonometric identities and other methods for quite some time with no results.
    Thank you
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  2. #2
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    Quote Originally Posted by karfag View Post
    Hello everyone,
    I've been preparing for my cal exam and I got stuck on this problem:
    Evaluate the infinite integral
    \int \frac {\sec^4 x}{\sqrt(tan 2x)} dx
    I would really appreciate if you give me a clue on how I should do it because I've been trying different trigonometric identities and other methods for quite some time with no results.
    Thank you
    Midread the integrand...
    Last edited by AllanCuz; May 23rd 2010 at 04:55 PM.
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  3. #3
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    I think you did not notice that it is a square root of tan(2x)... That is exactly what I was doing except that with square root it does not seem to lead anywhere.
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  4. #4
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    Quote Originally Posted by karfag View Post
    I think you did not notice that it is a square root of tan(2x)... That is exactly what I was doing except that with square root it does not seem to lead anywhere.
    Edit- lag. You're right, give me a moment i'll try something
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  5. #5
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    Hi everyone,

    I don't think we can solve this integral using general methods. I mean it needs some advanced mathematics. Please refer, Wolfram Mathematica Online Integrator

    Hope this will help you.
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  6. #6
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    Quote Originally Posted by Sudharaka View Post
    Hi everyone,

    I don't think we can solve this integral using general methods. I mean it needs some advanced mathematics. Please refer, Wolfram Mathematica Online Integrator

    Hope this will help you.
    Holy bleeping ****, yeah i give up right now LOL
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  7. #7
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    Eh, well I found it in previous final exam. First year college calculus.. I know the answer but I wanted to know how to get it. Well thanks for a try, if I'll figure it out, I'll post it lol
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  8. #8
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    Quote Originally Posted by karfag View Post
    Eh, well I found it in previous final exam. First year college calculus.. I know the answer but I wanted to know how to get it. Well thanks for a try, if I'll figure it out, I'll post it lol
    Dear karfag,

    Did your answer tally with the answer "Wolfram" suggested??
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  9. #9
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    The answer listed is

    (tan(2x))^1/2 + 1/5 * (tan(2x))^5/2 + C

    Sorry for not writing it the correct way, I'm not very familiar with the math tags syntax
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  10. #10
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    Quote Originally Posted by karfag View Post
    The answer listed is

    (tan(2x))^1/2 + 1/5 * (tan(2x))^5/2 + C

    Sorry for not writing it the correct way, I'm not very familiar with the math tags syntax
    Dear karfag,

    Differentiate (tan(2x))^{1/2} + \frac{1}{5} (tan(2x))^{5/2}+C and you would obtain, \frac {\sec^{4} 2x}{\sqrt{tan 2x}}. Hence the integration should be \int\frac {\sec^{4} 2x}{\sqrt{tan 2x}}
    Last edited by Sudharaka; May 24th 2010 at 11:42 PM. Reason: Wrong differentiation.
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  11. #11
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    If we let u = \sqrt{\tan{2x}}, then by the chain rule: \dfrac{du}{dx} = \dfrac{\sec^2{2x}}{\sqrt{\tan{2x}}} \Rightarrow {dx} = \dfrac{\sqrt{\tan{2x}}}{\sec^2{2x}}\;{du}. Thus \int\dfrac{\sec^4{x}}{\sqrt{\tan{2x}}}\;{dx} = \int\left(\dfrac{\sec^4{x}}{\sqrt{\tan{2x}}}\right  )\left(\dfrac{\sqrt{\tan{2x}}}{\sec^2{2x}}\right)\  ;{du} = \int\dfrac{\sec^4{x}}{\sec^2{2x}}\;{du}. It gets bit complicated after that, though.
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  12. #12
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    ... and since \sec^4(x) = \left(1+\tan^2{x}\right)\left(1+\tan^2{x}\right) and  \sec^2{2x} = \left\{\dfrac{1+\tan^2{x}}{1-\tan^2{x}}\right\}^2, then [tex]\dfrac{\sec^4{x}}{\sec^2{2x}}[/Math]  = \dfrac{\left(1+\tan^2{x}\right)\left(1+\tan^2{x}\r  ight)}{\left\{\dfrac{1+\tan^2{x}}{1-\tan^2{x}}\right\}^2}  = \dfrac{\left(1+\tan^2{x}\right)^2\left(1-\tan^2{x}\right)^2}{\left(1+\tan^2{x}\right)^2} = \left(1-\tan^2{x}\right)^2 = 1-2\tan^2{x}+\tan^4{x}. Therefore \int\dfrac{\sec^4{x}}{\sec^2{2x}}\;{du} = \sqrt{\tan{2x}}-2\int{\tan^2{x}}\;{du}+\int{\tan^4{x}}\;{du}<br />
.
    Last edited by TheCoffeeMachine; May 24th 2010 at 08:40 AM.
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  13. #13
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    Quote Originally Posted by TheCoffeeMachine View Post
    ... and since \sec^4(x) = \left(1+\tan^2{x}\right)\left(1+\tan^2{x}\right) and  \sec{2x} = \dfrac{1+\tan^2{x}}{1-\tan^2{x}}, then \color{red}{\dfrac{\sec^4{x}}{\sec{2x}} = \dfrac{\left(1+\tan{x}\right)\left(1+tan^2{x}\righ  t)}{\left(\dfrac{1+\tan^{x}}{1-\tan^2{x}}\right)}}  = \dfrac{\left(1+\tan^2{x}\right)\left(1+tan^2{x}\ri  ght)\left(1-\tan^2{x}\right)}{\left(1+\tan^2{x}\right)} = \left(1+\tan^2{x}\right) \left(1-\tan^2{x}\right) = 1-\tan^4{x}. Therefore \int\dfrac{\sec^4{x}}{\sec^2{2x}}\;{du} = \int{\left(1-\tan^4{x}\right)}\;{du} = u-\int{\tan^4{x}}\;{du} = \sqrt{\tan{2x}}-\int\tan^4{x}\;{du}.
    Dear TheCoffeeMachine,

    There are some incorrect calculations which I have highlighted. Please recheck them. It should be,

    \frac{\sec^{4}x}{\sec^{2}2x} = \frac{\left(1+\tan^{2}x\right)\left(1+\tan^{2}x\ri  ght)}{\left(\frac{1+\tan^{2}x}{1-\tan^{2}x}\right)^2}

    Hope this will help you.
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  14. #14
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    Quote Originally Posted by karfag View Post
    The answer listed is

    (tan(2x))^1/2 + 1/5 * (tan(2x))^5/2 + C

    Sorry for not writing it the correct way, I'm not very familiar with the math tags syntax

    The integrand should be  \frac{ \sec^4(2x) }{\sqrt{ \tan(2x) }} .

    If it is  \sec^4(x) , i believe that the integral cannot be expressed in terms of elementary function . I turn it into another one :

     \int \sqrt{ \frac{1}{x} - x }~dx but the definite integral

    \int_0^1 \sqrt{ \frac{1}{x} - x }~dx  I think it should be in terms of  \Gamma(\frac{1}{4}) .
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  15. #15
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    Quote Originally Posted by Sudharaka View Post
    There are some incorrect calculations which I have highlighted. Please recheck them.
    Fixed. Thanks. It's become: \sqrt{\tan{2x}}-2\int{\tan^2{x}}\;{du}+\int{\tan^4{x}}\;{du}.
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