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    Right well im stuck on a volume of revolution question, i have to integrate ((10cos(x-alpha))^2 but i dont know how to actually square that.
    Apparently the answer of ((10cos(x-alpha))^2 is 100*0.5(1+cos(2x-2alpha))

    any one able to show me how to get to that please?
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    Think about what you get if you do (2x)^2
    Apply the same to (10cos(x-a))^2
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    (Original post by Ben121)
    Think about what you get if you do (2x)^2
    Apply the same to (10cos(x-a))^2
    That does not help!!
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    It should do
    (2x)^2 is 4x^2. You square both parts inside.
    Doing that with your expression gives you (100 * cos(x-a)^2)
    You can then use an identity to rewrite cos(x-a)^a and integrate.
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    (Original post by Ben121)
    It should do
    (2x)^2 is 4x^2. You square both parts inside.
    Doing that with your expression gives you (100 * cos(x-a)^2)
    You can then use an identity to rewrite cos(x-a)^a and integrate.
    Ow yeah sorry i get that, my question should of been i dont get which identity to use then.
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    (Original post by Sheldon)
    Right well im stuck on a volume of revolution question, i have to integrate ((10cos(x-alpha))^2 but i dont know how to actually square that.
    Apparently the answer of ((10cos(x-alpha))^2 is 100*0.5(1+cos(2x-2alpha))

    any one able to show me how to get to that please?
    you've put 3 open brackets and only 2 close brackets. ??
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    (cos x)^2 = 0.5(1 + cos(2x))
 
 
 
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