# How do you integrate #int tsin(2t)# by integration by parts method?

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To integrate ( t \sin(2t) ) using integration by parts, you can apply the following formula:

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

Let ( u = t ) and ( dv = \sin(2t) , dt ), then differentiate ( u ) to get ( du ), and integrate ( dv ) to get ( v ):

[ du = dt ] [ v = -\frac{1}{2} \cos(2t) ]

Now, apply the integration by parts formula:

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

Integrate the second term on the right side:

[ \int -\frac{1}{2} \cos(2t) , dt = -\frac{1}{4} \sin(2t) ]

Therefore, the integral becomes:

[ \int t \sin(2t) , dt = -\frac{1}{2} t \cos(2t) + \frac{1}{4} \sin(2t) + C ]

Where ( C ) is the constant of integration.

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