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How do you integrate #int xsin(2x)# by integration by parts method?

Answer 1

Evelyn Agard

#=1/4sin(2x) - x/2cos(2x) + C#

For #u(x), v(x)#

#int uv'dx = uv ' - int u'vdx#

#u(x) = x implies u'(x) = 1#

#v'(x) = sin(2x) implies v(x) = -1/2cos(2x)#

#intxsin(2x)dx = -x/2cos(2x) +1/2intcos(2x)dx#

#= -x/2cos(2x)+1/4sin(2x)+C#

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Answer 2

Isabella Ackerman

To integrate ( \int x \sin(2x) ) using integration by parts, we use the formula:

[ \int u , dv = uv - \int v , du ]

Let ( u = x ) and ( dv = \sin(2x) , dx ).

Then, differentiate ( u ) to get ( du ) and integrate ( dv ) to get ( v ):

[ du = dx ] [ v = -\frac{1}{2} \cos(2x) ]

Now, apply the integration by parts formula:

[ \int x \sin(2x) , dx = -\frac{1}{2} x \cos(2x) - \int (-\frac{1}{2} \cos(2x)) , dx ]

[ = -\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) + C ]

Where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.