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Trouble with integral

  1. Jul 3, 2006 #1
    Hi, Ive been having trouble with this one its probly easy, but I cant get started.

    [tex]\int sinxcos(cosx)[/tex]

    I tried integrating by parts,

    du=-sin(cosx)(-sinx)......is this even right?
    but it looks as if doing that made it alot harder, can anyone point me in the right direction?thanks
  2. jcsd
  3. Jul 3, 2006 #2


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    Did you try a substitution?
  4. Jul 3, 2006 #3
    [tex]-\int cosu du[/tex]
    [tex]=-sin u[/tex]

    I pray this is wrong because if it isnt i'm retarted
  5. Jul 3, 2006 #4


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    That's it. (plus a constant, +C)
  6. Jul 3, 2006 #5
    I have another question which isnt explained in the text, when trying to find out A B and C in a partial fraction integral, how do you do it, I read online somewhere about gaussian elimination, it wasnt well explained though, is this how, or is there another way, can you explain?

  7. Jul 3, 2006 #6


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    Gaussian elimination???

    I assume that by "A B and C in a partial fraction integral" you mean, for example, the A, B, C is
    [tex]\frac{1}{(x-1)(x-2)(x-3)= \frac{A}{x-1}+ \frac{B}{x-2}+ \frac{C}{x-2}[/tex]

    Multiply the entire equation by the denominator (x-1)(x-2)(x-3) and you get
    [tex]1= A(x-2)(x-3)+ B(x-1)(x-2)+ C(x-1)(x-2)[/tex]

    Now let x= 1, 2, 3 in succession and A, B, C fall out. If you denominators are (x-a)2 or x2+ 1 you might need to let x equal other numbers to get additional equations to solve.
  8. Jul 3, 2006 #7


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    There's more than one way to approach it. An example:


    The denominator is distinct linear factors, so we know we will have:


    for some A and B. this becomes:


    Next equate the numerators:


    the 'placeholder' 0x is so you see the linear term is zero on the left.

    Equating coefficients you now have a system of 2 equations and 2 variables:


    Gaussian elimination is a general method for solving linear systems like this, but you can use whatever you like. Here, A=-B from the first equation, then the second becomes 1=2*A, so A=1/2 and B=-1/2.

    Rather than collect terms when you combined the fractions on the right hand side above, you could equate the numerators as:


    and sub in some convenient values for x. x=1 gives 1=A*(2)+B*(0) so A=1/2. x=-1 gives 1=A*(0)+B*(-2), so B=-1/2.
  9. Jul 3, 2006 #8
    i understand what you are saying..heres what im trying to figure out...

  10. Jul 3, 2006 #9


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    Well you can use the first equation to write B in terms of A, giving B=1-A.

    Substitute this value of B into the 2nd and 3rd equations and you now have 2 equations with 2 unknowns (A and C). Solve either equation for either A or C and substitute into the remaining equation and you'll have one of the variables. Can you see how to do this and where to go from here?
  11. Jul 3, 2006 #10
    yea i think im getting it...Il see if i can get an answer!
  12. Jul 3, 2006 #11
    ok, that wasnt too tough, i think i got it
  13. Jul 3, 2006 #12
    one final question..if i have [tex] \frac{x^3}{(x+3)^3}[/tex]..would it be this...[tex]\frac{A}{x+3} + \frac{B}{(x+3)^2} + \frac {C}{(x+3)^3}[/tex]..or would i have A B C and D...[tex]\frac{A}{1} + \frac{B}{x+3}[/tex]and so on?
  14. Jul 3, 2006 #13


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    In that case you have to do polynomial division first to get the numerator in lower terms than the denominator, and then it's the 3 term decomposition.
  15. Jul 3, 2006 #14
    oooooone more..heres my question..[tex]\int ln(x^2-1)dx[/tex]..i let..
    so....i kept going and got [tex]ln(x^2-1)x - \int \frac{2x^2}{x^2-1}dx[/tex]..would this partial fraction be A/x+1 +(B/x-1)?if so im confused because i get A+B=0 and A-B=0
    Last edited: Jul 3, 2006
  16. Jul 3, 2006 #15
    If u is ln(x^2+1) then du is definitely not 2x. What is the derivative of the natural log?
  17. Jul 3, 2006 #16
    yeah my bad..i meant to say [tex]2x \frac{1}{x^2-1}[/tex]
  18. Jul 3, 2006 #17
    Before you try decomposing into partial fractions you should divide 2x2 by x2-1 and that should give you something pluse a linear term over x2-1.
  19. Jul 3, 2006 #18
    good thinkin, I thinkI got it, THanks leet
  20. Jul 3, 2006 #19
    Your welcome, glad I could help.
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