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Math Help - Integral

  1. #1
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    Integral

    Hello All

    I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

    Bobak

    Edit: the integral is \int \frac{\sin^2((k+1)x)}{\sin^2x}
    Attached Thumbnails Attached Thumbnails Integral-picture-13.png  
    Last edited by bobak; June 20th 2008 at 10:49 AM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by bobak View Post
    Hello All

    I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

    Bobak
    Hello,

    I am sorry, I hate this notation. Is this

    \sin^2(k+1)\cdot{x}

    Or

    \sin^2((k+1)x)
    ?
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Hello,

    I am sorry, I hate this notation. Is this

    \sin^2(k+1)\cdot{x}

    Or

    \sin^2((k+1)x)
    ?
    the second one
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  4. #4
    Moo
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    Hello

    Quote Originally Posted by bobak View Post
    Hello All

    I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

    Bobak
    yay ! I think I found it ! ^^

    \frac{sin[(2k+1)x]}{sin(x)}=\frac{sin(2kx)cos(x)+cos(2kx)sin(x)}{sin  (x)}

    =\frac{2cos(kx)sin(kx)cos(x)+sin(x)[cos^2(kx)-sin^2(kx)]}{sin(x)}

    Multiply above and below by sin(x) :

    =... (skipped the step... I guess you can do it )

    Rearranging :

    =\frac{2{\color{red}sin(x)cos(kx)} \cdot {\color{blue}sin(kx)cos(x)}+[{\color{red}sin(x)cos(kx)}]^2-[sin(x)sin(kx)]^2}{sin^2(x)}

    Complete the square in red :

    =\frac{\left(\overbrace{{\color{red}sin(x)cos(kx)}  +{\color{blue}sin(kx)cos(x)}}^{sin[(k+1)x]}\right)^2-[{\color{blue}sin(kx)cos(x)}]^2-[sin(x)sin(kx)]^2}{sin^2(x)}


    \frac{sin[(2k+1)x]}{sin(x)}=\frac{sin^2[(k+1)x]}{sin^2(x)}-\frac{[sin(kx)cos(x)]^2+[sin(x)sin(kx)]^2}{sin^2(x)}


    ----------------------------

    \frac{[sin(kx)cos(x)]^2+[sin(x)sin(kx)]^2}{sin^2(x)}

    =\frac{sin^2(kx)cos^2(x)+sin^2(x)sin^2(kx)}{sin^2(  x)}=\frac{sin^2(kx)[cos^2(x)+sin^2(x)]}{sin^2(x)}=\boxed{\frac{sin^2(kx)}{sin^2(x)}}

    ----------------------------

    Integrating :

    -------> \frac{\pi}{2}=\overbrace{\int_0^{\pi /2} \frac{sin[(2k+1)x]}{sin(x)} \ dx}^{\frac{\pi}{2}}=\int_0^{\pi /2} \frac{sin^2[(k+1)x]}{sin^2(x)} \ dx - \int_0^{\pi /2} \frac{sin^2(kx)}{sin^2(x)} \ dx \quad \square
    Last edited by Moo; June 20th 2008 at 11:19 AM. Reason: multi edit for displaying :)
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  5. #5
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    Thanks for that Moo. Much apperiaceted.

    The last part is pretty straightforward

    Bobak
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  6. #6
    Moo
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    Quote Originally Posted by bobak View Post
    Thanks for that Moo. Much apperiaceted.

    The last part is pretty straightforward

    Bobak
    I edited many times my post, so I'm not sure what you're talking about
    Refresh ^^
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  7. #7
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    Quote Originally Posted by Moo View Post
    I edited many times my post, so I'm not sure what you're talking about
    Refresh ^^
    Yeah I noticed you editing it, I got the general idea thought, I made the error of trying to work form the result backwards, Which didn't work out very well for me.

    Bobak
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  8. #8
    Super Member PaulRS's Avatar
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    Quote Originally Posted by bobak View Post
    Hello All

    I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

    Bobak

    Edit: the integral is \int \frac{\sin^2((k+1)x)}{\sin^2x}
    In case you want to see an optional solution to the integral in part b) (not much to do with that exercise)

    <br />
I_n  = \int_{ - \pi }^\pi  {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}<br />
{{\sin \left( x \right)}}dx}  = 2\pi ,\forall n \in \mathbb{N}<br />

    By Euler's Formula: <br />
\sin \left( {n \cdot x} \right) = \tfrac{{e^{i \cdot n \cdot x}  - e^{ - i \cdot n \cdot x} }}<br />
{{2 \cdot i}};\sin \left( {n \cdot x} \right) = \tfrac{{e^{i \cdot x}  - e^{ - i \cdot x} }}<br />
{{2 \cdot i}}<br />
(we're not using contours)

    Thus <br />
I_n  = \int_{ -\pi }^{\pi } {\frac{{e^{i \cdot <br />
{\left( {2n + 1} \right)}<br />
 \cdot x}  - e^{ - i \cdot <br />
{\left( {2n + 1} \right)}<br />
 \cdot x} }}<br />
{{e^{i \cdot x}  - e^{ - i \cdot x} }}dx} <br />

    Now recall the formula: <br />
x^n  - y^n  = \left( {x - y} \right) \cdot \sum\limits_{k = 0}^{n - 1} {x^k  \cdot y^{n - 1 - k} } <br />
we have: <br />
e^{i \cdot \left( {2n + 1} \right) \cdot x}  - e^{ - i \cdot \left( {2n + 1} \right) \cdot x}  = \left( {e^{i \cdot x}  - e^{ - i \cdot x} } \right) \cdot \sum\limits_{k = 0}^{2n} {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} } <br /> <br />

    Thus: <br />
I_n  = \int_{ - \pi }^\pi  {\left( {\sum\limits_{k = 0}^{2n} {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} } } \right)dx}  = \sum\limits_{k = 0}^{2n} {\int_{ - \pi }^\pi  {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} } <br /> <br />

    Note that: <br />
\left\{ \begin{gathered}<br />
  {\text{If }}k \ne n \Rightarrow \int_{ - \pi }^\pi  {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx}  = 0 \hfill \\<br />
  {\text{If }}k = n \Rightarrow \int_{ - \pi }^\pi  {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx}  = 2\pi  \hfill \\ <br />
\end{gathered}  \right.<br />
Therefore: <br />
I_n  = \sum\limits_{k = 0}^{2n} {\int_{ - \pi }^\pi  {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} }  = 2\pi <br />

    And since: <br />
\int_{ - \pi }^\pi  {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}<br />
{{\sin \left( x \right)}}dx}  = 4 \cdot \int_0^{\tfrac{\pi }<br />
{2}} {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}<br />
{{\sin \left( x \right)}}dx} <br />
we have: <br />
\int_0^{\tfrac{\pi }<br />
{2}} {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}<br />
{{\sin \left( x \right)}}dx}  = \tfrac{\pi }<br />
{2},\forall n \in \mathbb{N}<br />
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  9. #9
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    I worked out I_{n}=\int_{0}^{\pi}\frac{sin^{2}(nx)}{sin^{2}(x)}  dx

    if you still care.

    For all positive integer values of n.

    Let's show that I_{n}=n{\pi}, \;\ n=1,2,3,....

    We have for n\geq{2}:

    D_{n}=I_{n}-I_{n-1}=\int_{0}^{\pi}\frac{(sin^{2}(nx)-sin^{2}((n-1)x)}{sin^{2}(x)}dx

    =\int_{0}^{\pi}\frac{(sin(nx)-sin((n-1)x))(sin(nx)+sin((n-1)x))}{sin^{2}(x)}dx

    \int_{0}^{\pi}\frac{2sin(\frac{x}{2})cos(n-\frac{x}{2})2sin(nx-\frac{x}{2})cos(\frac{x}{2})}{sin^{2}(x)}dx

    =\int_{0}^{\pi}\frac{sin(x)sin((2n-1)x)}{sin^{2}(x)}dx

    =\int_{0}^{\pi}\frac{sin((2n-1)x)}{sin(x)}dx

    So, D_{n}=J_{2n-1}, \;\ n\geq{2}

    J_{m}=\int_{0}^{\pi}\frac{sin(mx)}{sin(x)}dx, \;\ m=0,1,2,..

    Now, for m\geq{2}:

    J_{m}-J_{m-2}=\int_{0}^{\pi}\frac{sin(mx)-sin((m-2)x)}{sin(x)}dx

    =\int_{0}^{\pi}\frac{2sin(x)cos((m-1)x)}{sin(x)}dx

    =2\int_{0}^{\pi}cos((m-1)x)dx

    =2\left[\frac{sin((m-1)x)}{m-1}\right]_{0}^{\pi}=0

    So, we have

    J_{m}=J_{m-2}=J_{m-4}=....=\begin{gathered}J_{0}=0, \;\ \text{if m even}\\ J_{1}={\pi}, \;\ \text{if m odd}\end{gathered}

    Therefore, hence and hereftofore:

    We get D_{n}={\pi}, \;\ n\geq{2}, so that I_{n}=n{\pi}, \;\ n\geq{1}, \;\ as \;\ I_{1}={\pi}

    WHEW!!

    I just noticed. I used Pi instead of Pi/2 all this time. Oh well. Alas.
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  10. #10
    Moo
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    Hi galactus... Isn't it too complicated ?

    \int_0^{\pi /2} \frac{\sin^2 (kx)}{\sin^2(x)} \ dx=k \frac{\pi}{2}

    ---------------------
    For k=0, \frac{\sin^2(kx)}{\sin^2(x)}=0 \quad \square.

    Assume that \int_0^{\pi /2} \frac{\sin^2(kx)}{\sin^2(x)} \ dx = k \frac{\pi}{2}

    From the previous question :

    \int_0^{\pi /2} \frac{\sin^2 [(k+1)x]}{\sin^2(x)} \ dx=\int_0^{\pi /2} \frac{\sin^2 (kx)}{\sin^2(x)} \ dx + \frac{\pi}{2}=\bf (k+1) \frac{\pi}{2} \quad \blacksquare.

    huh?
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