Results 1 to 6 of 6

Math Help - sin(x)*sin(ax)

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    8

    Question sin(x)*sin(ax)

    Hi everyone,

    Is there a trig identity for combining
     <br />
sin(x)*sin(ax)<br />

    and if not, what's the best way to integrate this?

    Thanks,
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by brian311 View Post
    Hi everyone,

    Is there a trig identity for combining
     <br />
sin(x)*sin(ax)<br />

    and if not, what's the best way to integrate this?

    Thanks,
    No there is no identity that I know of, the solution is to integrate by parts twice and realise that the second integral is equal to the question being asked. Also you should be very careful with your + and - signs

    I = \int \sin(x) \sin(ax)

    u = \sin(ax) \: \rightarrow \: u' = a \cos(ax)

    v' = \sin(x) \: \rightarrow \: v = -\cos(x)

    I = -\sin(ax) \cos(x) + a\int \cos(x) \cos(ax)


    Integrate by parts on the integral we have on the RHS

    u = \cos(ax) \: \rightarrow \: u' = -a \sin(ax)

    v' = \cos(x) \: \rightarrow \: v = \sin(x)


    \int \cos(x) \cos(ax) = (\sin(x) \cos(ax) + a\int \sin(x) \sin(ax))


    Putting this into the original integration by parts.

    I = -\sin(ax) \cos(x) + a(\sin(x) \cos(ax) + a\int \sin(x) \sin(ax))

    Note that we have the question, I, here.

    I = -\sin(ax) \cos(x) + a(\sin(x) \cos(ax) + aI) = -\sin(ax) \cos(x) + a\sin(x) \cos(x) + a^2I

    I(1-a^2) = -\sin(ax) \cos(x) + a\sin(x) \cos(x)

    Isolate I and this expression will be equal to the question
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
     \sin x \sin ax = \frac{1}{2} \Big(\cos (1-a)x - \cos (1+a)x \Big)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2008
    Posts
    8
    Quote Originally Posted by e^(i*pi) View Post
    No there is no identity that I know of, the solution is to integrate by parts twice and realise that the second integral is equal to the question being asked. Also you should be very careful with your + and - signs

    I = \int \sin(x) \sin(ax)

    u = \sin(ax) \: \rightarrow \: u' = a \cos(ax)

    v' = \sin(x) \: \rightarrow \: v = -\cos(x)

    I = -\sin(ax) \cos(x) + a\int \cos(x) \cos(ax)


    Integrate by parts on the integral we have on the RHS

    u = \cos(ax) \: \rightarrow \: u' = -a \sin(ax)

    v' = \cos(x) \: \rightarrow \: v = \sin(x)


    \int \cos(x) \cos(ax) = (\sin(x) \cos(ax) + a\int \sin(x) \sin(ax))


    Putting this into the original integration by parts.

    I = -\sin(ax) \cos(x) + a(\sin(x) \cos(ax) + a\int \sin(x) \sin(ax))

    Note that we have the question, I, here.

    I = -\sin(ax) \cos(x) + a(\sin(x) \cos(ax) + aI) = -\sin(ax) \cos(x) + a\sin(x) \cos(x) + a^2I

    I(1-a^2) = -\sin(ax) \cos(x) + a\sin(x) \cos(x)

    Isolate I and this expression will be equal to the question
    Thanks for your help!

    I think there may have been a small typo, I'm getting:
    I(1-a^2) = -\sin(ax) \cos(x) + a\sin(x) \cos(ax)

    instead of:
    I(1-a^2) = -\sin(ax) \cos(x) + a\sin(x) \cos(x)

    Thanks again!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by brian311 View Post
    Thanks for your help!

    I think there may have been a small typo, I'm getting:
    I(1-a^2) = -\sin(ax) \cos(x) + a\sin(x) \cos(ax)

    instead of:
    I(1-a^2) = -\sin(ax) \cos(x) + a\sin(x) \cos(x)

    Thanks again!
    You're right, it's hard to keep track of without seeing it infront of me on a sheet of paper
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    You don't have to integrate by parts.

     \int  \sin x \sin ax = \frac{1}{2} \int \Big(\cos (1-a)x - \cos (1+a)x \Big) \ dx


     =  \frac{\sin(1-a)x}{2(1-a) } - \frac{\sin (1+a)x}{2(1+a)} +C

    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum