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Math Help - Looking for an integration pointer!

  1. #1
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    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 u-substitution path I can generally work my way through, but I'm hitting a block on these, even just a pointer or what u should be would be great!

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

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

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

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

    Any toss as to where to go next would be great, thanks for taking a look!
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  2. #2
    Super Member Random Variable's Avatar
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     \sin^{2}u = \frac {1}{2} - \frac {\cos 2u}{2}
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  3. #3
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    Quote Originally Posted by superduper View Post

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

    For this one, I'm trying to find a u sub but I can't seem to make it work, I keep wishing real hard that the x^2 was under the radical and the x was outside.
    Put t=\sqrt{x+1} written backwards as t^2=x+1.
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  4. #4
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    Hey, so I got an answer for the radical integrand, but I'm still having trouble with the Trig question.

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

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

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

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

    EDIT- I'm mistaken, what you posted was 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.
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  5. #5
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    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.
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  6. #6
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    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

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

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

    So they say

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

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

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

    Thanks for your time!
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  7. #7
    Super Member Random Variable's Avatar
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    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  \sin^{2} \Big(\frac{u}{2}\Big) = \frac {1-\cos u}{2}

     \cos^{2} \Big(\frac {u}{2} \Big) = \frac {1+\cos u}{2}
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  8. #8
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    But the original equation was sin(3x) so if we make u=2x don't we have to turn that into \frac{3}{2}u=3x or something to make that equation usable? Or actually \frac{2}{3}u=\frac{2}{3}(3x)? I'm getting lost now.
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  9. #9
    Super Member Random Variable's Avatar
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    Quote Originally Posted by superduper View Post
    But the original equation was sin(3x) so if we make u=2x don't we have to turn that into \frac{3}{2}u=3x or something to make that equation usable? Or actually \frac{2}{3}u=\frac{2}{3}(3x)? I'm getting lost now.
     \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.
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  10. #10
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    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?

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

    Where u=3x and 3du=dx

    Therefore,

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

    Looking good so far? Thanks again for your time.
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  11. #11
    Super Member Random Variable's Avatar
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    if you let u = 3x , then  du=3dx which implies that  \frac {du}{3} = dx

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

    Now you need to change the limits of integration.

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

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

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

    now using the identity

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

     \frac {u}{6} \Big|^{\pi}_{0} - \frac{\sin(2u)}{12} \Big|^{\pi}_{0}
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  12. #12
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    Perfect, I can finish with that. Just one quick question on the last step.

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

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

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

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

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

    Thanks!
    You have to make a substituion.

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

    so the intergal becomes  \frac {1}{2} \int \cos v \ dv = \frac {1}{2} \sin v = \frac {1}{2} \sin(2u)
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  14. #14
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    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.
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