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∫ xsin(2x) dx = (-1/2)xcos2x + (1/4)sin2x
You get this by using Integration by Parts.
An integral in the form ∫udv can be written as uv-∫vdu
In the case of your problem u=x, du=1, dv=sin2x, v=(-1/2)cos2x <--You get v by integrating dv
Using the formula ∫udv = uv- ∫vdu and by plugging in what has been defined above you get ∫xsin(2x)dx = (-1/2)xcos2x - ∫(-1/2)cos2x(1)
By integrating ∫(-1/2)cos2x, you get (-1/4)sin2x.
When you plug that back in, you get ∫xsin2xdx=(-1/2)xcos2x-(-1/4)sin2x or just simply
∫xsin(2x)dx = (-1/2)xcos(2x) + (1/4)sin(2x)
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8
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What is the indefinite integral of 3sinx-5cosx?
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Integral Systems was created in 1982.
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