How do you integrate #int t sin(mt)dt# where m does not equal 0?

Answer 1

#(sin(mt)-mtcos(mt))/m^2+C#

We have:

#inttsin(mt)dt#, where #m# is presumably real, and does not equal zero. We say that #m!=0# and #m inRR#.
According to integration by parts, #intf(x)g(x)dx=f(x)intg(x)dx-intf'(x)(intg(x)dx)dx#
Here, #f(t)=t# and #g(t)=sin(mt)#. We have:
#tintsin(mt)dt-int(intsin(mt)dt)dt#
We need to compute #intsin(mt)dt#. Using integration by substitution,
#intf(g(x))g'(x)dx=intf(u)du#, where #u=g(x)#. Here, #u=mt#, and #u'=m#, so we can say:
#1/m intsin(mt)mdt#
#1/m intsin(u)du#
#1/m*-cos(u)#
#-cos(mt)/m#. Inputting:
#-(tcos(mt))/m-int-cos(mt)/mdt#
#-(tcos(mt))/m+1/m intcos(mt)dt#
Using integration by substitution, where #u=mt# again, we have:
#-(tcos(mt))/m+1/m(1/m intcos(mt)mdt)#
#-(tcos(mt))/m+1/m(1/m intcos(u)du)#
#-(tcos(mt))/m+1/m(1/m sin(mt))#
#sin(mt)/m^2-(tcos(mt))/m#
#(sin(mt)-mtcos(mt))/m^2+C#
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Answer 2

To integrate ∫t sin(mt) dt, where m ≠ 0, you can use integration by parts. Let u = t and dv = sin(mt) dt. Then differentiate u to get du = dt, and integrate dv to get v = -cos(mt)/m. Apply the integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values of u, v, du, and dv into the formula:

∫t sin(mt) dt = -t(cos(mt))/m - ∫(-cos(mt)/m) dt

Now, integrate the second term:

∫(-cos(mt)/m) dt = (1/m) ∫cos(mt) dt

Using the fact that the integral of cosine is sine:

= (1/m) * (-sin(mt)/m) + C

Combine terms:

= -t(cos(mt))/m + sin(mt)/(m^2) + C

So, the integral of ∫t sin(mt) dt is -t(cos(mt))/m + sin(mt)/(m^2) + 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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