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    This is the question

    Integral[(sin 2t)(cos t) dt]

    I tried doing this numerous times but I can never get rid of the multiplication between two functions. Its always goes to (sin 2t)(cos t) or (sin t)(cos 2t -1) which multiplies to get (sin t.cos 2t - sin t)
    See, I can't get rid of the multiplication so how can I integrate it?

    Any help appreciated!
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    http://en.wikipedia.org/wiki/List_of...uct_identities
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    Not sure if this is right... but
    Convert Sin2t = 2SintCost
    and then see what you can do that way.
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    (Original post by Without a Shadow)
    EDIT: Im wrong
    'Fraid so

    OP, try using integration by substitution.

    Convert it using the double angle formula as the others have said, then use substitution.

    u = cos x so du/dx = -sin x.

    See it?
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    There are two approaches here:

    1) The laborious way. If you integrate by parts twice, then the integrand you are left with will be in the same form as the original one, so you should be able to do it with a bit of algebraic manipulation.

    2) The elegant and quick way (and possibly the way they are getting at in the question). Rewrite the sin 2t using your trusty double-angle formula. Then see if you can notice that the integrand happens to be a perfect derivative of something due to the chain rule.

    Hope that helps, I've put the answer I got as a spoiler for you:

    Spoiler:
    Show
    -2/3 * cos^3 (t)
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    (Original post by sphixter)
    This is the question

    Integral[(sin 2t)(cos t) dt]

    I tried doing this numerous times but I can never get rid of the multiplication between two functions. Its always goes to (sin 2t)(cos t) or (sin t)(cos 2t -1) which multiplies to get (sin t.cos 2t - sin t)
    See, I can't get rid of the multiplication so how can I integrate it?

    Any help appreciated!
    sin(2t+t) = sin2tcost + cos2tsint
    sin(2t-t) = sin2tcost - cos2tsint

    Add these together:

    sin3t + sint = 2sin2tcost

    sin2tcost = 1/2(sin3t + sint)

    SAFE.

    EDIT, oh sorry then integrating it is piss easy. -1/6cos3t - 1/2cost
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    (Original post by rob_the_sax_dude)
    Hope that helps, I've put the answer I got as a spoiler for you:
    Hi, i got something different, so i wonder if you would be able to tell me where i went wrong?

    Spoiler:
    Show

    \int (sin2t)(cost) \, dt 



= \int (2sintcost)(cost) \, dt



= \int (2sintcos^2t) \, dt



= \int (2sint(1-sin^2t) \, dt



= \int (2sint - 2sin^3t) \, dt



= -2cost + 1/2cost^4t + c
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    (Original post by _bonnie)
    Hi, i got something different, so i wonder if you would be able to tell me where i went wrong?

    [SPOILER]
    \int (sin2t)(cost) \, dt 



= \int (2sintcost)(cost) \, dt



= \int (2sintcos^2t) \, dt



= \int (2sint(1-sin^2t) \, dt



= \int (2sint - 2sin^3t) \, dt



= -2cost + 1/2cost^4t + c





e.g. [latex]\sin^2 \theta + \cos^2 \theta=1
    Most probably the same thing.

    If you see my solution above, it is also different.
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    (Original post by _bonnie)
    Hi, i got something different, so i wonder if you would be able to tell me where i went wrong?

    Spoiler:
    Show

    \int (sin2t)(cost) \, dt 



= \int (2sintcost)(cost) \, dt



= \int (2sintcos^2t) \, dt



= \int (2sint(1-sin^2t) \, dt



= \int (2sint - 2sin^3t) \, dt



= -2cost + 1/2cost^4t + c
    = \int (2sint - 2sin^3t) \, dt



= -2cost + 1/2cost^4t + c

    that is what i disagree with, you cant integrate powers like that. instead leave the fiorst line as sin x cos^2 x, thensubstitute u = cos x.
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    of course you can't, opps!
    thanks for your help
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    (Original post by aster100)
    sin(2t+t) = sin2tcost + cos2tsint
    sin(2t-t) = sin2tcost - cos2tsint

    Add these together:

    sin3t + sint = 2sin2tcost

    sin2tcost = 1/2(sin3t + sint)

    SAFE.

    EDIT, oh sorry then integrating it is piss easy. -1/6cos3t - 1/2cost
    Thanks! Can't believe I didn't think of that.
 
 
 
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