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Math Help - I have a questions!!

  1. #1
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    I have a questions!!

    INT [(x+Sinx)/(1+cosx)]dx
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by berkanatci View Post
    INT [(x+Sinx)/(1+cosx)]dx
    Divide this into two integrals:
    Int[(x+Sinx)/(1+cosx) dx] = Int[x/(1 + cos(x)) dx] + Int[sin(x)/(1 + cos(x)) dx]

    The first integral can be done by parts:
    Int[p dq] = pq - Int[q dp]

    Let p = x, dp = dx
    Let dq = dx/(1 + cos(x)), q = Int[dx/(1 + cos(x))] = tan(x/2)

    So
    Int[x/(1 + cos(x)) dx] = x*tan(x/2) - Int[tan(x/2) dx] = x*tan(x/2) - (-2)*ln{|cos(x/2)|} = x*tan(x/2) + 2*ln{|cos(x/2)|}

    The second integral may be done by substitution: y = cos(x), dy = -sin(x) dx:
    Int[sin(x)/(1 + cos(x)) dx] = -Int[dy/(1 + y)] = -ln{|y + 1|} = -ln{|cos(x) + 1|}

    Thus:
    Int[(x+Sinx)/(1+cosx) dx] = x*tan(x/2) + 2*ln{|cos(x/2)|} - ln{|cos(x) + 1|}

    Now, |cos(x/2)| = sqrt{(1 + cos(x))/2} so
    ln{|cos(x/2)|} = ln[sqrt{(1 + cos(x))/2}] = ln[sqrt{(1 + cos(x))}] - ln[sqrt{2}] = (1/2)ln[cos(x) + 1] - (1/2)ln[2] so.....

    Int[(x+Sinx)/(1+cosx) dx] = x*tan(x/2) + 2*ln{|cos(x/2)|} - ln{|cos(x) + 1|}

    = x*tan(x/2) + ln{cos(x) + 1} - ln{2} - ln{|cos(x) + 1|}

    = x*tan(x/2) - ln{2}

    and, of course we need to add an arbitrary constant, so I'd say that

    Int[(x+Sinx)/(1+cosx) dx] = x*tan(x/2) + C

    -Dan
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by berkanatci View Post
    INT [(x+Sinx)/(1+cosx)]dx
    That x is not very easy to deal with. If you have a typo such that that x is actually supposed to be a 1, then this integration will be easy, but with an x I think it will take a few by-parts integrations to deal with.

    INT (x + sinx)/(1 + cosx) dx

    Let's multiply the numerator and denominator by 1 - cosx
    INT [(x + sinx)(1 - cosx)]/(1 - cos^2x) dx
    INT (x - x*cosx + sinx - sinxcosx)/sin^2x dx

    Let's break this up into 4 fractions:
    INT x/sin^2x dx - INT x*cosx/sin^2x dx + INT sinx/sin^2x - INT sinx*cosx/sin^2x dx
    INT x*csc^2x dx - INT x*cotx*cscx dx + INT cscx dx - INT cotx dx

    Now we will integrate each of these individually:

    First:
    INT x*csc^2x dx

    Let u = x <--> du = dx
    Let dv = csc^2x dx <--> v = -cotx
    INT x*csc^2x dx = -xcotx + INT cotx dx = -xcotx + ln(abs(sinx))

    Second:
    INT x*cotx*cscx dx

    Let u = x <--> du = dx
    Let dv = cotx*cscx dx <--> v = -cscx
    INT x*cotx*cscx dx = -x*cscx + INT cscx dx = -x*cscx - ln(abs(cscx + cotx))

    Third:
    INT cscx dx = -ln(abs(cscx + cotx))

    Forth:
    INT cotx dx = ln(abs(sinx))

    Therefore:
    INT (x + sinx)/(1 + cosx) dx = (-xcotx + ln(abs(sinx))) - (-x*cscx - ln(abs(cscx + cotx))) + (-ln(abs(cscx + cotx))) - (ln(abs(sinx))) + C

    I don't feel like simplifying that ... Check to make sure I did the integrations correctly. I suspect I may have made a few mistakes.

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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by topsquark View Post
    Divide this into two integrals:
    Int[(x+Sinx)/(1+cosx) dx] = Int[x/(1 + cos(x)) dx] + Int[sin(x)/(1 + cos(x)) dx]

    The first integral can be done by parts:
    Int[p dq] = pq - Int[q dp]

    Let p = x, dp = dx
    Let dq = dx/(1 + cos(x)), q = Int[dx/(1 + cos(x))] = tan(x/2)

    So
    Int[x/(1 + cos(x)) dx] = x*tan(x/2) - Int[tan(x/2) dx] = x*tan(x/2) - (-2)*ln{|cos(x/2)|} = x*tan(x/2) + 2*ln{|cos(x/2)|}

    The second integral may be done by substitution: y = cos(x), dy = -sin(x) dx:
    Int[sin(x)/(1 + cos(x)) dx] = -Int[dy/(1 + y)] = -ln{|y + 1|} = -ln{|cos(x) + 1|}

    Thus:
    Int[(x+Sinx)/(1+cosx) dx] = x*tan(x/2) + 2*ln{|cos(x/2)|} - ln{|cos(x) + 1|}

    Now, |cos(x/2)| = sqrt{(1 + cos(x))/2} so
    ln{|cos(x/2)|} = ln[sqrt{(1 + cos(x))/2}] = ln[sqrt{(1 + cos(x))}] - ln[sqrt{2}] = (1/2)ln[cos(x) + 1] - (1/2)ln[2] so.....

    Int[(x+Sinx)/(1+cosx) dx] = x*tan(x/2) + 2*ln{|cos(x/2)|} - ln{|cos(x) + 1|}

    = x*tan(x/2) + ln{cos(x) + 1} - ln{2} - ln{|cos(x) + 1|}

    = x*tan(x/2) - ln{2}

    and, of course we need to add an arbitrary constant, so I'd say that

    Int[(x+Sinx)/(1+cosx) dx] = x*tan(x/2) + C

    -Dan
    Though mine's not simplified, do you think my answer is equivolent to yours? I didn't think to use a half-angle formula for:
    Int[dx/(1 + cos(x))] = tan(x/2)
    How did you get that, by the way?
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post

    INT (x + sinx)/(1 + cosx) dx = (-xcotx + ln(abs(sinx))) - (-x*cscx - ln(abs(cscx + cotx))) + (-ln(abs(cscx + cotx))) - (ln(abs(sinx))) + C
    I'll simplify it, I'm curious:
    - x*cotx + ln(abs(sinx)) + x*cscx + ln(abs(cscx + cotx)) - ln(abs(cscx + cotx)) - ln(abs(sinx)) + C

    = x*cscx - x*cotx + C
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  6. #6
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    Quote Originally Posted by ecMathGeek View Post
    Though mine's not simplified, do you think my answer is equivolent to yours? I didn't think to use a half-angle formula for:


    How did you get that, by the way?


    thank u.now I understand better than the past....

    but I have many ınt problems ıf u try to solve them,I ll be very happy...
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    Though mine's not simplified, do you think my answer is equivolent to yours? I didn't think to use a half-angle formula for:


    How did you get that, by the way?
    Quote Originally Posted by ecMathGeek View Post
    I'll simplify it, I'm curious:
    - x*cotx + ln(abs(sinx)) + x*cscx + ln(abs(cscx + cotx)) - ln(abs(cscx + cotx)) - ln(abs(sinx)) + C

    = x*cscx - x*cotx + C
    (sigh) To be honest I did it on my calculator and it looked like a simple enough result I never went back to see if I could do it by hand. I CAN'T (baby-like wail). So I'll simplify your result down to mine:
    csc(x) - cot(x) = 1/sin(x) - cos(x)/sin(x) = (1 - cos(x))/sin(x)

    = (1 - cos(x))/sqrt(1 - cos^2(x)) = (1 - cos(x))/[sqrt(1 - cos(x))*sqrt(1 + cos(x))]

    = sqrt(1 - cos(x))/sqrt(1 + cos(x)) = sqrt((1 - cos(x))/2)/sqrt((1 + cos(x))/2)

    = sin(x/2)/cos(x/2) = tan(x/2)

    -Dan
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