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Math Help - help meeeeee:(:(:(((

  1. #1
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    help meeeeee:(:(:(((

    1)INT dx/(cosx+sinx)^4 ????



    2)INT dx/(sin^3x+cos^3x) ????



    3)INT dx/(sinx+2secx)^2 ?????



    4)INT dx/[(sin^4x)cosx] ????


    5)INT [(cosx + sinx)/(sin2x^1/2)]dx ??????
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by berkanatci View Post
    1)INT dx/(cosx+sinx)^4 ????



    2)INT dx/(sin^3x+cos^3x) ????



    3)INT dx/(sinx+2secx)^2 ?????



    4)INT dx/[(sin^4x)cosx] ????


    5)INT [(cosx + sinx)/(sin2x^1/2)]dx ??????
    INT 1/(cosx + sinx)^4 dx

    Multiply out the denominator:
    INT 1/(cos^4x + 4cos^3x*sinx + 6cos^2x*sin^2x + 4cosx*sin^3x + sin^4x) dx

    Now, group terms to reduce:
    INT 1/[(cos^4x + 2cos^2x*sin^2x + sin^4x) + (4cos^3x*sinx + 4cosx*sin^3x) + 4cos^2x*sin^2x] dx
    INT 1/[(cos^2x + sin^2x)^2 + 4cosx*sinx(cos^2x + sin^2x) + 4cos^2x*sin^2x] dx
    INT 1/(1 + 4cosx*sinx + 4cos^2x*sin^2x) dx

    When factored, the denominator becomes:
    INT 1/(2cosx*sinx +1)^2 dx
    INT 1/(sin2x + 1)^2 dx

    Multiply numerator and denominator by (1 - sin2x)^2
    INT (1 - sin2x)^2/(1 - sin^2(2x))^2 dx
    INT (1 - 2sin2x + sin^2(2x))/cos^4(2x) dx

    Let 1 = sin^2(2x) + cos^2(2x)
    INT (cos^2(2x) - 2sin2x + 2sin^2(2x))/cos^4(2x) dx

    Separate the fraction:
    INT cos^2(2x)/cos^4(2x) dx - INT 2sin(2x)/cos^4(2x) dx + INT 2sin^2(2x)/cos^4(2x) dx
    INT sec^2(2x) dx - 2*INT sec^2(2x)[tan2x*sec2x] dx + 2*INT tan^2(2x)[sec^2(2x)] dx

    First:
    INT sec^2(2x) dx = 1/2*tan2x

    Second:
    INT sec^2(2x)[tan2x*sec2x] dx

    Let u = sec2x <--> du = 2*tan2x*sec2x dx --> dx = 1/(2tan2x*sec2x) du
    1/2*INT u^2 du
    1/6*u^3
    1/6*sec^3(2x)

    Third:
    INT tan^2(2x)[sec^2(2x)] dx

    Let u = tan2x <--> du = 2*sec^2(2x) dx --> dx = 1/(2sec^2(2x)) du
    1/2*INT u^2 du
    1/6*u^3
    1/6*tan^3(2x)

    Putting these back to gether, we get:
    (
    1/2*tan2x) - 2(1/6*sec^3(2x)) + 2(1/6*tan^3(2x)) + C

    1/2*tan2x - 1/3*sec^3(2x) + 1/3*tan^3(2x) + C

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  3. #3
    Senior Member ecMathGeek's Avatar
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    OMG, I would be making some good money if I were actually paid to do this stuff.
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  4. #4
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    These questions so easy berkan.
    You must work too much...
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by pinsim View Post
    These questions so easy berkan.
    I wouldn't say that

    Look at http://www.mathhelpforum.com/math-he...ease-help.html

    At least, the second one is very easy, rather than berkanatci's problems.
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