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Homework Help: Some calc questions

  1. Feb 13, 2006 #1
    Hi, just needa clear up how to do some certain questions before i start uni in a few weeks.

    Intergrate: sin(x) . cos^4 (x)

    I'm not sure if i needa split up the coses to use an identity or whatever.

    intergrate: xe^(x^2+1)

    = xe(x^2+1)/2x
    (1/2)e(x^2+1) ?

    I think that is right but not certain.


    Intergrate: 5/m - 3√m expressing answer in surd form

    I get as far as 5ln|m| .. then get slightly confused by the second part.

    Cheers in advance to thoose who can help and shows steps to refresh me.
  2. jcsd
  3. Feb 14, 2006 #2
    Well the first one you're gonna need to know some of those annoying trig substitutions. I haven't had to memorize them in years, but they're the formulas for like sin^m*cos^n when m is odd and n even, or both odd, and so on. There're specific ways to deal with that

    The second one is integration by parts, which I HAVE had to memorize since cal 2, so you should make note to know it too. If you have the integral of u*dv, that can be expressed as u*v-int(v*du) it's a sorta circular exchange of variables which makes it easy to memorize. As for which is u and which is dv, you should experiment, remembering you want the integral you end up with in the second equation to actually be solvable, and remember that the dv includes the (dx) you didn't mention, so when you integrate to find v, you know you're integrating with respect to x.

    as for 5/m - 3/m, I can't tell if that's like a square root or something you were trying to make there. If so, writing it as 3(m)^(1/2) should make it more clear
  4. Feb 14, 2006 #3

    That second integral does NOT require integration by parts a simple u-substitution will work.
  5. Feb 14, 2006 #4


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    Homework Helper

    For 1,
    If the power of sine function is odd, then just make a u-substitution: u = cos(x).
    [tex]\int \sin ^ 3 x \cos ^ 2 x dx[/tex]. Let u = cos x => du = -sin x dx (or dx = du / (-sin x)). Now, we have:
    [tex]\int \sin ^ 3 x \cos ^ 2 x dx = \int -\sin ^ 2 x \times u ^ 2 du = \int (u ^ 2 - 1) u ^ 2 du = \int (u ^ 4 - u ^ 2) du = \frac{u ^ 5}{5} - \frac{u ^ 3}{3} + C[/tex]
    [tex]= \frac{\cos ^ 5 x}{5} - \frac{\cos ^ 3 x}{3} + C[/tex].
    If the power of cosine function is odd, then just make a u-substitution: u = sin(x).
    For 2, hint:
    [tex]3 \sqrt{m} = 3 m ^ \frac{1}{2}[/tex].
    For 3, d_leet has pointed out that this can be done with a u-substitution, what's u, do you think?
    Can you go from here? :)
  6. Feb 14, 2006 #5
    Oh, well, yah, of course. I was...you know...giving him..practice..at..IBP. Yah. Practice. Because it's useful. And good...to practice >_>
  7. Feb 14, 2006 #6
    Great i've got the 2nd and 3rd question, but the first one seems to be a gap in my studies, i have never heard of this odd function concept, could someone please work through the question with labeled steps for me as it may refresh me.
  8. Feb 14, 2006 #7
    When you integrate (sinx)^m*(cosx)^n, where m and n are just integers(don't misread my notation, it's the (sinx)^m, not sin(x^2))you can do it by making a u substitution that isn't necessarily obvious on inspection. As someone mentioned, when m is odd, you do u=cos(x)(I'm assuming he was right)in your problem m is 1, so yah, u=cosx, du=-sin(x), and it's a u substitution, which you know how to do if you can do the second problem, you may just have not studied that specific case and then you'd have to sorta guess on u substitutions until you got something that worked.
  9. Feb 14, 2006 #8
    seriously I've tried this a few times now and the answer they give is still different, can someone please take the time to do the question for me :)

    i have reviewed viet's technique and what he has done with the sin^3 x is really confusing me.
  10. Feb 15, 2006 #9
    Integrand is sin(x)*cos^4(x), make the sub u=cos(x) so du=-sin(x)dx

    if u=cos(x), you transform the integrand into -u^4*du (you almost had the du there(the sin(x)) but you needed that negative so you just multiply your new integrand by -1/-1, which is of course just multiplying by one so it's a legal operation, and it gets you the negative sin(x) to make your du, but you still had that -1 you were dividing by, which then makes it negative...err, it's hard to explain and I'm assuming you've studied making these substitutions before, as you did the other problem fine apparently)

    the integral of -u^4*du is -(u^5)/5. Substitute cos(x) back in for u, so the answer is -(cos^5(x))/5
    Last edited: Feb 15, 2006
  11. Feb 15, 2006 #10
    The part that lost you in his example, I believe where after he peels off one of those sines(so up to there it works like your problem, but in the end he has that extra sin^2(x) left over)he made the trig sub sin^2(x)=1-cos^2(x), or 1-u^2, and he had that negative there from making the du(just like I did)so it becomes u^2-1
  12. Feb 15, 2006 #11
    finally got it, ~_~!

    Cheers the help on this one schatten~
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