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

Answer 1

#I = -t cos(2t)/2 + 1/4 sin(2t)#

Set #sin(2t)dt = dv# then #v = -cos(2t)/2# #u = t# #du = dt#
Using #int udv = [uv] - int v du# now you get #I = -t cos(2t)/2 + 1/2 int cos(2t) dt# #I = -t cos(2t)/2 + 1/4 sin(2t)#
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Answer 2

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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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.

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.

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.

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.

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