Results 1 to 10 of 10

Math Help - Techniques of integration (3)

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Techniques of integration (3)

    Let n \geq 0 be an integer. Evaluate I_n=\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Using itegration by parts I could probably kick this thing off...


    \int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx = \frac{1}{n}cos^n(x)sin(nx) + \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx

    Not really sure how to do,

     \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx

    probably needs some trig identities to tidy it up.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by pickslides View Post
    Using itegration by parts I could probably kick this thing off...


    \int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx = \color{red}\frac{1}{n}cos^n(x)sin(nx)  + \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx.

    Not really sure how to do  \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx.

    probably needs some trig identities to tidy it up.
    that's a good start. we have I_0=\frac{\pi}{2}. for n \geq 1, as you (almost) showed, we have: I_n=\int_0^{\frac{\pi}{2}} \sin(nx) \sin(x) \cos^{n-1}x \ dx. now what?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Okay, I think I got the correct idea

    From \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b), it follows that :
    \sin(nx)\sin(x)=\cos[(n-1)x]-\cos(nx)\cos(x)

    The integral is thus :

    I_n=\int_0^{\frac\pi2} \cos[(n-1)x]\cos^{n-1}(x)-\cos(nx)\cos(x)\cos^{n-1}(x) ~dx

    I_n=\int_0^{\frac\pi2} \cos[(n-1)x]\cos^{n-1}(x) ~dx-\int_0^{\frac\pi2} \cos(nx)\cos^n(x) ~dx

    Which is :
    I_n=I_{n-1}-I_n

    I_n=\frac 12 \cdot I_{n-1}

    \Rightarrow \boxed{I_n=\frac{\pi}{2^{n+1}}}

    That's a really nice one
    Last edited by Moo; May 15th 2009 at 12:20 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Bravo Moo! that's exactly the idea ... although you made a strange mistake at the end of your solution, which i'm sure you'll easily find and fix it!
    Last edited by NonCommAlg; May 15th 2009 at 01:56 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by NonCommAlg View Post
    Brave Moo! that's exactly the idea ... although you made a strange mistake at the end of your solution, which i'm sure you'll easily find and fix it!
    Yes indeed

    1+1=n+1 : that's a very well-known result !!
    Had a too short night...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Consider
     cos^n(x) = \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} [cos(n-2k)x + i sin(n-2k)x]<br />


    Since  cos^n(x) is real , the terms of sine function will be cancelled finally ,

    but we experience that  \int_0^{\frac\pi2} cos(n-2k)x ~cosnx ~dx = 0 if n-2k is not equal to n so the integral becomes :

     \frac{\pi}{2^{n+1}}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by simplependulum View Post

     \cos^n(x) = \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} [\cos(n-2k)x + i \sin(n-2k)x]<br />
    this is not an identity!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by NonCommAlg View Post
    Let n \geq 0 be an integer. Evaluate I_n=\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx.
    It can be done using complex variables...

    First note that

    4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\int_0^{2\pi} (\cos x)^n \cos (nx) \ dx

    Parameterize the unit circle in the comples plane with

    z=e^{i\theta} \implies \frac{dz}{iz}=d\theta

    So the integral becomes

    \oint \left(\frac{z+z^{-1}}{2}\right)^n\left( \frac{z^n+z^{-n}}{2}\right)\frac{dz}{iz}

    expanding with the binomial theorem we get

    \frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{n-i}z^{-i}\right]\left[z^{n-1}+z^{-n-1} \right]dz

    \frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{2n-2i-1}+ \sum_{i=0}^{n}\binom{n}{i}z^{i-1}\right]

    each finite sume is its own larent series so the residues are coeffients on the \frac{1}{z} terms so we get

    \frac{1}{i\cdot 2^{n+1}}\left[2\pi i (1)+2\pi i (1) \right]=\frac{4\pi}{2^{n+1}}

    4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{4\pi}{2^{n+1}}

    \int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{\pi}{2^{n+1}}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by TheEmptySet View Post
    It can be done using complex variables...

    First note that

    4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\int_0^{2\pi} (\cos x)^n \cos (nx) \ dx

    Parameterize the unit circle in the comples plane with

    z=e^{i\theta} \implies \frac{dz}{iz}=d\theta

    So the integral becomes

    \oint \left(\frac{z+z^{-1}}{2}\right)^n\left( \frac{z^n+z^{-n}}{2}\right)\frac{dz}{iz}

    expanding with the binomial theorem we get

    \frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{n-i}z^{-i}\right]\left[z^{n-1}+z^{-n-1} \right]dz


    \frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{2n-2i-1}+ \sum_{i=0}^{n}\binom{n}{i}z^{i-1}\right]
    i think the power of z in the second sum should be -2i-1.


    each finite sume is its own larent series so the residues are coeffients on the \frac{1}{z} terms so we get

    \frac{1}{i\cdot 2^{n+1}}\left[2\pi i (1)+2\pi i (1) \right]=\frac{4\pi}{2^{n+1}}

    4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{4\pi}{2^{n+1}}

    \int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{\pi}{2^{n+1}}
    looks good to me!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Techniques of integration (6)
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: May 22nd 2009, 05:54 PM
  2. Techniques of integration (5)
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: May 20th 2009, 06:09 AM
  3. Techniques of integration (4)
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: May 18th 2009, 03:58 AM
  4. Techniques of integration (2)
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: May 2nd 2009, 06:09 AM
  5. Techniques of integration (1)
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: April 24th 2009, 06:50 PM

Search Tags


/mathhelpforum @mathhelpforum