The integral of (sin 3t)^5 cos t dt

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Discussion Overview

The discussion revolves around the integral of (sin 3t)^5 cos t dt, exploring methods for solving it, including u-substitution and potential expansions. Participants share their approaches and reasoning, highlighting the complexity of the integral.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using u-substitution but struggles with the setup.
  • Another participant proposes setting u = sin(3t) and expresses confusion about the disappearance of cos(t) in the conversion.
  • A different approach is mentioned involving the expansion of sin(3t) using deMoivre's theorem, suggesting it may lead to a more complex integral.
  • Some participants note that the integral is complicated due to the multiplication of a power of sine with a cosine of a different argument.
  • One participant provides a detailed expansion of sin(3t) and proposes a substitution with u = sin(t), leading to a new integral form.
  • Another participant describes the integral as easy but messy, indicating that while the method is known, the execution is lengthy.
  • A later reply gives a specific result for the integral, presenting a complex expression involving multiple cosine terms.
  • One participant mentions the need to apply the binomial formula for further evaluation after substitution.

Areas of Agreement / Disagreement

Participants express varying methods and approaches to the integral, with no consensus on a single solution or method. The discussion remains unresolved with multiple competing views on how to tackle the integral.

Contextual Notes

Some participants acknowledge the complexity and potential length of the solution process, indicating that assumptions about the ease of integration may not hold due to the integral's structure.

PhysicsMajor
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Hey folks,

HERE GOES... :confused:

the integral of (sin 3t)^5 cos t dt

i believe you have to use u substitution but i am having trouble getting it set up correctly.

Thanks for any input

:smile:
 
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[tex]\int{sin(3t)^5}cos(t)}{dt}[/tex]

u = sin(3t)
du = 3cos(3t) dt

[tex]1/3\int{u^5*3cos(3t)}{dt}[/tex]

[tex]1/3\int{u^5}{du}[/tex]
 
whozum said:
[tex]\int{sin(3t)^5}cos(t)}{dt}[/tex]

u = sin(3t)
du = 3cos(3t) dt

[tex]1/3\int{u^5*3cos(3t)}{dt}[/tex]

[tex]1/3\int{u^5}{du}[/tex]

How does that work? What happened to the cos t?
 
From converting the integral to one with respect to u instead of t. The conversion factor is right above the integral
 
Ugh I did it wrong hold on. I am so tired
 
PhysicsMajor said:
Hey folks,

HERE GOES... :confused:

the integral of (sin 3t)^5 cos t dt

i believe you have to use u substitution but i am having trouble getting it set up correctly.

Thanks for any input

:smile:
There may be easier ways, but you could expand:

[tex]\sin(3t)=4sin(t)cos(t)^2-sin(t)[/tex]
by exploiting deMoivre's theorem.
 
Mathworld's Integrator gives a long and not-very-nice answer btw.
 
Then you'd have to raise that to the fifth and foil and blah blah blah. Its a real complicated integral.
 
Well, you ARE multiplying a power of a sin with a cosine with a different argument, I don't expect the answer to be nice.
Besides, you just have to develop a 'work-attitude' in some situations. Roll up your sleeves and do it. It may be the fastest way. It won't take more than a few minutes, while looking for a possible easier way probably takes longer.
 
  • #10
[tex]sin(3t) = sin(2t+t) = sin(2t)cos(t) + sin(t)cos(2t) = 2sin(t)cos^2(t) + sin(t)(cos^2(t)-sin^2(t))[/tex]

[tex]= 2sin(t)(1-sin^2(t)) + sin(t)(1-2sin^2(t)) = 2sin(t)-2sin^3(t) + sin(t)-2sin^3(t) = 3sin(t)-4sin^3(t)[/tex]



[tex]\int{sin(3t)^5cos(t)}{dt}[/tex]

[tex]\int{(3sin(t)-5sin^3(t))^5cos(t)}{dt}[/tex]

[tex]u=sin(t), du = cos(t) dt[/tex]

[tex]\int{(3u-4u^3)^5}{du}[/tex]

I think that's as good as it gets if I didnt make any errors.
 
Last edited:
  • #11
This is integral is the typical example of an easy & but messy integral.Eas,because you know what to do to get to the result and messy,because it would take a page of writing to do it...

Daniel.

P.S.[tex]\sin 3x=-4\sin^{3}x+3\sin x[/tex]
 
  • #12
[tex]\int \left( \left( \sin 3x\right) ^5\cos x\right) dx= -\frac 1{512}\cos 16x-\frac 1{448}\cos 14x+\frac 1{64}\cos 10x+ \frac 5{256}\cos 8x-\frac 5{64}\cos 4x-\frac 5{32}\cos 2x +C[/tex]

Daniel.
 
  • #13
Making the substitution

[tex]\sin x=u[/tex]

u'll need to evaluate this cutie pie

[tex]\int \left(-4u^3+3u\right)^{5} \ du[/tex]

So use the binomial formula.

Daniel.
 

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