Results 1 to 14 of 14

Thread: Looking for an integration pointer!

  1. #1
    Newbie
    Joined
    Jul 2009
    Posts
    18

    Looking for an integration pointer!

    Hey guys, I got two problems here, that are probably extremely basic; but i'm having a hard time with these. When I finally find a $\displaystyle u$-substitution path I can generally work my way through, but I'm hitting a block on these, even just a pointer or what $\displaystyle u$ should be would be great!

    1) $\displaystyle \int x^2 \sqrt{x+1}dx$

    For this one, I'm trying to find a $\displaystyle u$ sub but I can't seem to make it work, I keep wishing real hard that the $\displaystyle x^2$ was under the radical and the $\displaystyle x$ was outside.

    2) $\displaystyle \int\sin^2{(3x)}dx$
    $\displaystyle = \int(\sin{(3x)})^2dx$

    Again, no idea what to set $\displaystyle u$ equal to because there is no $\displaystyle 2(\sin{(3x)})(\cos{(3x)})(3)$ with the $\displaystyle dx$

    Any toss as to where to go next would be great, thanks for taking a look!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    $\displaystyle \sin^{2}u = \frac {1}{2} - \frac {\cos 2u}{2} $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Quote Originally Posted by superduper View Post

    1) $\displaystyle \int x^2 \sqrt{x+1}dx$

    For this one, I'm trying to find a $\displaystyle u$ sub but I can't seem to make it work, I keep wishing real hard that the $\displaystyle x^2$ was under the radical and the $\displaystyle x$ was outside.
    Put $\displaystyle t=\sqrt{x+1}$ written backwards as $\displaystyle t^2=x+1.$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2009
    Posts
    18
    Hey, so I got an answer for the radical integrand, but I'm still having trouble with the Trig question.

    I set $\displaystyle u=3x$, and to integrate my textbook has a formula that states $\displaystyle \int\sin^2{u}du=\frac{1}{2}u-\frac{1}{4}\sin{2u}+C$.

    So I applied this as follows,
    $\displaystyle
    \int_{0}^{\frac{\pi}{3}}\sin^2{(3x)}dx$
    Where $\displaystyle u=3x$ and $\displaystyle du=3dx$

    Therefore,
    $\displaystyle
    \int_{0}^{\frac{\pi}{3}}\sin^2{(3x)}dx$
    = $\displaystyle 3\int_{0}^{\frac{\pi}{3}}\sin^2{(u)}du$
    =$\displaystyle 3(\frac{1}{2}u+\frac{1}{4}\sin{(2u)})|^{\pi/3}_{0}$
    =$\displaystyle [\frac{3}{2}\sin{((3)(\frac{\pi}{3}))}+\frac{3}{4}\ sin{((2)(3)(\frac{\pi}{3})}]-[\frac{3}{2}\sin{(0)}+\frac{3}{4}\sin{(0)}]$
    =$\displaystyle [\frac{3}{2}\sin{(\pi)}+\frac{3}{4}\sin{(2\pi)}]-[\frac{3}{2}\sin{(0)}+\frac{3}{4}\sin{(0)}]$
    =$\displaystyle [0+0+-[0+0]$?

    I'm not sure where I'm going wrong here.

    I was going to sub in the post that RandomVariable made, but I was just wondering if you could clarify that for me, I thought that identity was supposed to look like

    $\displaystyle sin^2{\frac{u}{2}}=\frac{1+cos(u)}{2}$

    EDIT- I'm mistaken, what you posted was $\displaystyle cos2u$ but I'm still lost as to where that identity comes from? I can't find it anywhere in my textbook.

    Thanks again for the help guys.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1
    Go to this website W|A
    Type in this exact phrase: integrate x^2 sqrt[x+1].
    Click the equals bar at the end if the input window.
    Be sure you click ‘show steps’ to see the solution.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2009
    Posts
    18
    Wow Plato, that website is amazing, thanks for showing me that!

    That solution showed me that I was wrong, but I was on the right track, messed up the u's somewhere.

    Right off the bat, WolframAlpha says

    $\displaystyle \int x^2\sqrt{x+1}dx$

    Where $\displaystyle u=\sqrt{x+1}$ and $\displaystyle du=\frac{1}{2\sqrt{x+1}}$

    So they say

    $\displaystyle \int x^2\sqrt{x+1}dx$
    = $\displaystyle 2\int u^2 (u^2-1)^2 du$
    = $\displaystyle 2\int (u^6-2u^4+u^2)du$

    Could you elaborate on this for me please? From the above relation shouldn't $\displaystyle u^2=x+1$ not $\displaystyle u^2=x^2$ I have no idea how they got this, I came up with

    $\displaystyle \int_{0}^3 (u-1)^2\sqrt{u}du$ where $\displaystyle u=x+1$
    = $\displaystyle \int_{0}^3 (x^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}})du$

    Thanks for your time!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    I was going to sub in the post that RandomVariable made, but I was just wondering if you could clarify that for me, I thought that identity was supposed to look like



    EDIT- I'm mistaken, what you posted was but I'm still lost as to where that identity comes from? I can't find it anywhere in my textbook.
    Just let u= 2x. And it should be $\displaystyle \sin^{2} \Big(\frac{u}{2}\Big) = \frac {1-\cos u}{2} $

    $\displaystyle \cos^{2} \Big(\frac {u}{2} \Big) = \frac {1+\cos u}{2} $
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jul 2009
    Posts
    18
    But the original equation was $\displaystyle sin(3x)$ so if we make $\displaystyle u=2x$ don't we have to turn that into $\displaystyle \frac{3}{2}u=3x$ or something to make that equation usable? Or actually $\displaystyle \frac{2}{3}u=\frac{2}{3}(3x)$? I'm getting lost now.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by superduper View Post
    But the original equation was $\displaystyle sin(3x)$ so if we make $\displaystyle u=2x$ don't we have to turn that into $\displaystyle \frac{3}{2}u=3x$ or something to make that equation usable? Or actually $\displaystyle \frac{2}{3}u=\frac{2}{3}(3x)$? I'm getting lost now.
    $\displaystyle \sin^{2}(3x) = \frac {1 - \cos2(3x)}{2} = \frac {1-\cos(6x)}{2} $

    The u=2x thing was to show that the identity you remember is the same identity.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jul 2009
    Posts
    18
    Oooh, that makes sense. Thanks for the clarification!

    So here's what I have so far, would you mind poking over it and letting me know what you think?

    $\displaystyle \int_{0}^{\frac{\pi}{3}} sin^2(3x)dx$

    Where $\displaystyle u=3x$ and $\displaystyle 3du=dx$

    Therefore,

    $\displaystyle \int_{0}^{\frac{\pi}{3}} sin^2(3x)dx$
    = $\displaystyle 3\int_{0}^{\frac{\pi}{3}} sin^2u du$
    = $\displaystyle 3\int_{0}^{\frac{\pi}{3}} \frac{1-cos(2u)}{2} du$
    = $\displaystyle \frac{3}{2} \int_{0}^{\frac{\pi}{3}} 1-cos(2u) du$
    = $\displaystyle \frac{3}{2}(u-sin(2u))|_{0}^{\frac{\pi}{3}}$

    Looking good so far? Thanks again for your time.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    if you let $\displaystyle u = 3x $, then $\displaystyle du=3dx $ which implies that $\displaystyle \frac {du}{3} = dx $

    so the integral becomes $\displaystyle \frac {1}{3} \int \sin^2(u) \ du $

    Now you need to change the limits of integration.

    The upper limit, which was $\displaystyle x=\frac{\pi}{3}$, becomes $\displaystyle u =3*\frac{\pi}{3} = \pi$

    The lower limt, which was $\displaystyle x=0 $, becomes $\displaystyle u =3*0=0$

    so the integral now becomes $\displaystyle \frac{1}{3}\int^{\pi}_{0} \sin^{2}(u) \ du $

    now using the identity

    $\displaystyle = \frac{1}{3} \int^{\pi}_{0} \big(\frac {1}{2} - \frac{\cos(2u)}{2} \Big) du $ $\displaystyle = \frac{1}{6}\int^{\pi}_{0} du - \frac{1}{6} \int^{\pi}_{0} \cos(2u) \ du $

    $\displaystyle \frac {u}{6} \Big|^{\pi}_{0} - \frac{\sin(2u)}{12} \Big|^{\pi}_{0} $
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Jul 2009
    Posts
    18
    Perfect, I can finish with that. Just one quick question on the last step.

    Why is $\displaystyle \int cos(2u)du=\frac{sin(2u)}{2}$? I don't see that formula anywhere either, I thought it was;

    $\displaystyle \int cos (u) du= sin (u)$

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by superduper View Post
    Perfect, I can finish with that. Just one quick question on the last step.

    Why is $\displaystyle \int cos(2u)du=\frac{sin(2u)}{2}$? I don't see that formula anywhere either, I thought it was;

    $\displaystyle \int cos (u) du= sin (u)$

    Thanks!
    You have to make a substituion.

    Let $\displaystyle v = 2u $, then $\displaystyle dv=2du $ which implies $\displaystyle \frac {dv}{2}=du $

    so the intergal becomes $\displaystyle \frac {1}{2} \int \cos v \ dv = \frac {1}{2} \sin v = \frac {1}{2} \sin(2u) $
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Jul 2009
    Posts
    18
    Oh I see, that makes sense, I've never done more than 1 substitution in the same problem before. Thanks for sticking with me here, really appreciate it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Nov 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: Nov 2nd 2010, 04:57 AM
  3. Replies: 2
    Last Post: Feb 19th 2010, 10:55 AM
  4. Need pointer on getting started
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: Sep 12th 2009, 11:28 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum