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Math Help - Integration by parts and with trig

  1. #1
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    Integration by parts and with trig

    I have two problems I don't really know how to work through.
    The first is the integral from 0 to 1 of r^3/(sqr(4+r^2))
    I started by integrating by parts and eventually ended up with (4+r^2)^(7/2) so I know I'm doing something wrong.

    The other problem is: find the integral of cosPIx * cos4PIx
    I don't know where to begin here. Any help would be appreciated.
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  2. #2
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    For first one you don't need to integrate by parts. Simply sub u = 4 + r^2.

    For the second one, you can either integrate by parts twice and treat the integral as the unknown or you can use the product to sum identity:

    \cos{a}\cos{b} = \frac{1}{2}(\cos{(a-b)} + \cos{(a+b)})
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  3. #3
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    Quote Originally Posted by LINKJ6 View Post
    The first is the integral from 0 to 1 of r^3/(sqr(4+r^2))
    <br />
\int {\frac{{r^3 }}<br />
{{\sqrt {4 + r^2 } }}dr} <br />

    define: <br />
4 + r^2  = x^2  \Rightarrow rdr = xdx<br />

    Now I will do something wrong, but help when I subtitute dx

    <br />
\int {\frac{{r^3 }}<br />
{x} \cdot \frac{x}<br />
{r}} dx = \int {r^2 dx} <br />

    but <br />
r^2  = x^2  - 4<br />
then

    <br />
\int {x^2  - 4dx}  = \frac{{x^3 }}<br />
{3} - 4x + C<br />

    now restoring we obtain ordered


    Quote Originally Posted by LINKJ6 View Post
    The other problem is: find the integral of cosPIx * cos4PIx
    I don't know where to begin here. Any help would be appreciated.
    In this case you can integrating by parts, but whit identiti given for Chop Suey the solution is direct
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  4. #4
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    define:

    Now I will do something wrong, but help when I subtitute dx



    but then



    now restoring we obtain ordered
    I'm confused by this. Is it double substitution or something? I tried substituting u for 4+r^2 but there's no du present.
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  5. #5
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    Quote Originally Posted by LINKJ6 View Post
    I'm confused by this. Is it double substitution or something? I tried substituting u for 4+r^2 but there's no du present.
    Link's method is not double substitution and it works.

    If you use Chops' method,

    u = 4 + r^2 yields
    du = 2r \, dr.

    Note that \int \frac{r^3}{\sqrt{4 + r^2}} dr = \frac{1}{2} \int \frac{2r^3 \, dr}{\sqrt{4 + r^2}} = \frac{1}{2} \int \frac{r^2 \cdot 2r \, dr}{\sqrt{4 + r^2}}, so you can substitute du for 2r dr and u - 4 for r^2:

    \frac{1}{2} \int \frac{u - 4}{\sqrt{u}}du
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  6. #6
    Member Nacho's Avatar
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    Quote Originally Posted by LINKJ6 View Post
    I'm confused by this. Is it double substitution or something? I tried substituting u for 4+r^2 but there's no du present.
    How do icemanfan is the correct way, I show my way, because I think that is easier see the change of dr to dx
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  7. #7
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    For the second one, I get to the integral of 1/2((cosPIx*cos4PIx + sinPIx*sin4PIx) + (cosPIx*cos4PIx - sinPIx*sin4PIx))dx

    I'm pretty sure I use substitution but I don't know what I substitute. Am I allowed to set u=cosPIx du=-PIsinPIx and putting a 4 in front of u for the cos4PIx?
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  8. #8
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    Why did you expand the cosines using the addition identity? You're complicating things. Just integrate \cos{(-3\pi x)} = \cos{(3\pi x)} as you would integrate a cos(x).

    If the negative sign inside is throwing you off, recall that Cosine is an even function:

    \cos{(-x)} = \cos{(x)}
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  9. #9
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    There's no negative in the cos. I just don't know if I'm "allowed" to take the 4 out of cos(4PIx) and put it in front of the cos. [4cosPIx]
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  10. #10
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    No, you can't do that.

    \cos{(\pi x}\cos{(4\pi x)}

    = \frac{1}{2}(\cos{(\pi x - 4 \pi x)} + \cos{(\pi x + 4 \pi x)})

    = \frac{1}{2}(\cos{(-3 \pi x)} + \cos{(5 \pi x)})

    = \frac{1}{2}(\cos{(3 \pi x)}) + \frac{1}{2}(\cos{(5 \pi x)})

    Just integrate it directly!
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  11. #11
    Member Nacho's Avatar
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    Quote Originally Posted by LINKJ6 View Post
    There's no negative in the cos. I just don't know if I'm "allowed" to take the 4 out of cos(4PIx) and put it in front of the cos. [4cosPIx]
    no, always that you have this question, restitute whit numbers

    in fact:
    <br />
\cos \left( {4\pi x} \right) = \cos \left( {2\pi x + 2\pi x} \right) = \cos ^2 \left( {2\pi x} \right) - \sin ^2 \left( {2\pi x} \right)<br />
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  12. #12
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    Oh. I'm so sorry I read the question wrong. Thank you all very much for the help. I greatly appreciate it.
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