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Math Help - Integration with trigonometry

  1. #1
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    Integration with trigonometry

    Hi,

    Can someone please show me how to solve this, step by step?

    Integrate the following: sinxcosxcos2x

    Apparently the answer is 1/8(sin2x)^2 + C

    Does sinxcosx go outside of the integrand as a factor?
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  2. #2
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    Quote Originally Posted by CSG18 View Post
    Hi,

    Can someone please show me how to solve this, step by step?

    Integrate the following: sinxcosxcos2x

    Apparently the answer is 1/8(sin2x)^2 + C

    Does sinxcosx go outside of the integrand as a factor?

    only constants can be factored out of an integral ... you know that.
    \sin{x}\cos{x}\cos(2x) =

    \frac{1}{2} \cdot 2\sin{x}\cos{x}\cos(2x) =

    \frac{1}{2} \cdot \sin(2x)\cos(2x)

    substitution ... let u = \sin(2x)
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \sin{x}\cos{x}\cos(2x) =

    \frac{1}{2} \cdot 2\sin{x}\cos{x}\cos(2x) =

    \frac{1}{2} \cdot \sin(2x)\cos(2x)

    substitution ... let u = \sin(2x)
    Sorry I do not understand what you did...

    How did it turn into sin(2x)\cos(2x) ?

    I understand where the fraction came from though.
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  4. #4
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    Quote Originally Posted by CSG18 View Post
    Sorry I do not understand what you did...

    How did it turn into sin(2x)\cos(2x) ?

    I understand where the fraction came from though.
    Because sin(2x) = 2 sinx cosx

    If you dont know this, you need to look back at the topic double angle formula. Here it is:

    Double Angle

    The fraction is there because skeeter multiplied the term \sin{x}\cos{x}\cos(2x) by  \frac{2}{2} , which is equal to 1
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  5. #5
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    Quote Originally Posted by CSG18 View Post
    Hi,

    Can someone please show me how to solve this, step by step?

    Integrate the following: sinxcosxcos2x

    Apparently the answer is 1/8(sin2x)^2 + C

    Does sinxcosx go outside of the integrand as a factor?
    A function of the variabel NEVER "goes outside the integrand"! Only constants (with respect to the variable of integration) can be moved like that.

    Skeeter used the trig identity that sin(2x)= 2sin(x)cos(x) so that sin(x)cos(x)= (1/2)sin(2x).
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    A function of the variabel NEVER "goes outside the integrand"! Only constants (with respect to the variable of integration) can be moved like that.

    So why is it when you integrate 4cosx sin^3x dx, 4 cosx goes outside as a factor?
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  7. #7
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    Quote Originally Posted by CSG18 View Post
    So why is it when you integrate 4cosx sin^3x dx, 4 cosx goes outside as a factor?
    No because \cos(x) is not a constant. 4 can go out of the integrand but this is not the optimal way to solve it

    If you use skeeter's hint of u = \sin(2x) then du = 2\cos(2x)\,dx

    Therefore \int \cos(2x) \sin(2x)\,dx = \frac{1}{2}\int u\,du


    Note that I have taken a factor of 1/2 out of the integral - this is because 1/2 is a constant
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