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Homework Help: How do i integrate this?

  1. Nov 26, 2007 #1
    how do i integrate this??

    a) ∫cot^(2)x cscx dx?

    b) ∫[x^(3) + 2x^(2) + x]^(1/2) dx
  2. jcsd
  3. Nov 26, 2007 #2
    I can tell you that the first problem requires you to use a trig identity followed by integration by parts and a little bit of manipulation.
    The second one requires factoring, and then distributing and from there it should be quite easy.

    Show me what you have done so far so we can take it from there.
    Last edited: Nov 26, 2007
  4. Nov 26, 2007 #3
    i've try to use
    u=cotx n=2, but
    du=-csc^(2)x dx

    then, i use
    u=csc x n=1, but
    du=-csc x cotx dx

    after that i don't know what to do!! i'm stuck there!
    Last edited: Nov 26, 2007
  5. Nov 26, 2007 #4
    so step 1, integration by parts

    let u=cotx, du=cscxcotxdx

    now what do you get? and next step?
  6. Nov 26, 2007 #5
    hmmm...substitution would not be ideal here as none is a derivative of the other and the derivative of cot(x) is -csc^2(x) not -cos^2(x).

    What i rather you do is to use the trig identity that cot^2(x) = csc^2(x) - 1
    and distribute to get csc^3(x) - csc(x). It will take more integration techniques to break this down further as I had said.

    csc^3(x) needs to be integrated using integration by parts and then a simple rearrangement.

    csc(x) should be done by multiplying it by (cscx + cotx)/(csx + cotx) and then using the general log rule to integrate.

    Since i will have to leave for class, the second one;

    rocophysics' method is less tedious than the one i had pointed out so its best to go that way.

    Just factor an x out first then factorize what you are left with. Then simplify and distribute then use the necessary integration rules to integrate.

    Last edited: Nov 26, 2007
  7. Nov 27, 2007 #6
    pls xhexk if i answered it correctly..

    ∫[x^(3) + 2x^(2) + x]^(1/2) dx

    =∫{x[x^(2) + 2x +1]}^(1/2) dx

    =∫ [x^(1/2)][(x+1)^(2)]^(1/2) dx

    then the square root will be canceled out in [(x+1)^(2)]^(1/2)

    =∫ [x^(1/2)](x+1) dx

    =∫x^(1/2) * x dx + ∫ x^(1/2) dx

    =∫ x^(3/2) dx + ∫ x^(1/2) dx

    =(2/5) x^(5/2) + (2/3) x(3/2) + C

    is that the final answer??
  8. Nov 27, 2007 #7
    yup thats it right to the end mon. good job
  9. Nov 27, 2007 #8
    the the first equation.. i don't te final answer but i have your idea..
    let me try some..

    ∫cot^(2)x cscx dx

    =∫ [csc^(2)x -1] cscx dx

    =∫[csc^(3)x -cscx]dx

    =∫csc^(3)x dx - ∫cscx dx

    =∫:confused: - ∫cscx dx *(cscx-cotx)/(cscx-cotx)

    =∫:confused: - ∫ [csc^(2)x -cscx cotx] dx / (cscx cotx)

    =∫:confused: - ln (cscx - cotx) +C

    i dont know how to integrate cot^(3)x but i'm trying to figure it out
  10. Nov 27, 2007 #9

    Gib Z

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    Homework Helper

    Perhaps an easier way to do a) is:

    [tex] I = - \int \cot x ( - \cot x \csc x) dx[/tex].
    Using The Pythagorean Identities to express cot x in terms of csc x, and letting u= csc x

    [tex]I = \int \sqrt{u^2-1} du[/tex]

    That last integral is quite readily done by a hyperbolic trig substitution.
  11. Nov 27, 2007 #10


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    [tex]x^3+2x^2+x=x(x^2+2x+1)[/tex] factorise that quadratic again and...then put it back under the square root sign and see if anything occurs
  12. Dec 2, 2007 #11
    ah rock freak I already told rhey to do that, and he/she got the answer. its the other one thats giving the problem now.

    you would have to use integration by parts for csc^(3)x.

    u = cscx dv = csc^(2)x dx
    du = -cscxcotx dx v = -cotx

    int(csc^(3)x)dx = -cscxcotx - int(cscx(csc^(2)x - 1))dx as cot^(2)x = csc^(2)x -1

    2*int(csc^(3)x)dx = -cscxcotx + int(cscx)dx
    int(csc^(3)x)dx = -(1/2)cscxcotx + (1/2)int(cscx)dx

    and you can do int(cscx)dx as you did it earlier in the problem. that substitution method looks easier though. but this is the first thing that popped in mind.
  13. Dec 3, 2007 #12


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    Integration by parts doesn't always give you an easier time than if you use a substitution...sometimes the easiest way is the best way
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