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Math Help - Please HELP

  1. #1
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    Please HELP

    My Questions

    1. INT(tan 4^x) dx

    2. INT ( (3x+4)/(x^2+1)^2 ) dx

    thank you.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pinsim View Post

    1. INT(tan 4^x) dx
    is it tan(4x) or tan^4(x)

    i'll assume the latter since the first would be too easy


    int{tan^4(x)}dx
    = int{tan^2(x)*tan^2(x)}dx
    = int{(sec^2(x) - 1)*tan^2(x)}dx .......trig identity: 1 + tan^2(x) = sec^2(x)
    = int{tan^2(x)sec^2(x) - tan^2(x)}dx ...........i multiplied out
    = int{tan^2(x)sec^2(x)}dx - int{tan^2(x)}dx ..............split the integral in 2

    for the first: int{tan^2(x)sec^2(x)}dx, we proceed by substitution
    let u = tan(x)
    => du = sec^2(x) dx
    so our integral becomes
    int{u^2}du
    = (1/3)u^3
    = (1/3)tan^3(x)

    for the second part: int{tan^2(x)}dx
    = int{sec^2(x) - 1}dx .............the same identity
    = tan(x) - x

    so int{tan^2(x)sec^2(x)}dx - int{tan^2(x)}dx = (1/3)tan^3(x) - tan(x) + x + C
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    still working on the second problem...
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  4. #4
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    Questions

    1. tan 4^x dx ( x is a exponent of 4)
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pinsim View Post
    1. tan 4^x dx ( x is a exponent of 4)
    oh, darn it
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  6. #6
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    Quote Originally Posted by pinsim View Post
    My Questions

    1. INT(tan 4^x) dx

    2. INT ( (3x+4)/(x^2+1)^2 ) dx

    thank you.
    Well the first one has no primitive.

    But the second one has, jeje

    Just split the integral in two, then you must solve one by substitution and the another one by trigonometric substitution.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Well the first one has no primitive.

    But the second one has, jeje

    Just split the integral in two, then you must solve one by substitution and the another one by trigonometric substitution.
    yeah, i think the same for the first..and for the second, but i was running into problems for the trig substituion part
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  8. #8
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    yes i tried but can't solve it .
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  9. #9
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    Well, I've got a solution for de second one...



    Like I said, split in two integrals and solve one by substitution, and the another one by trigonometric substitution, it's not hard, just set x = tan u and then it's done...
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  10. #10
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    When you try the trig. substitution, you'll have to integrate cosē u, but instead to integrate by parts, remember that cosē u = (1 + cos 2u)/2, which is more easy to integrate.
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  11. #11
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    heyy budiess ,
    thanks first of all for the second Question.
    But can you help me for the first one..
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  12. #12
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    Quote Originally Posted by pinsim View Post
    heyy budiess ,
    thanks first of all for the second Question.
    But can you help me for the first one..
    as Krizalid said, the first one does not have a solution in terms of elementary functions (we don't think). are you sure it's not a typo?
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  13. #13
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    the problem is ;

    INT tan 4^x dx ( x is a exponent of 4)
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pinsim View Post
    the problem is ;

    INT tan 4^x dx ( x is a exponent of 4)
    yeah, it has no solution that we can compute at this point
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