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

  • #1

Main Question or Discussion Point

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:
 

Answers and Replies

  • #2
2,209
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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]
 
  • #3
551
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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?
 
  • #4
2,209
1
From converting the integral to one with respect to u instead of t. The conversion factor is right above the integral
 
  • #5
2,209
1
Ugh I did it wrong hold on. Im so tired
 
  • #6
Galileo
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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.
 
  • #7
551
1
Mathworld's Integrator gives a long and not-very-nice answer btw.
 
  • #8
2,209
1
Then you'd have to raise that to the fifth and foil and blah blah blah. Its a real complicated integral.
 
  • #9
Galileo
Science Advisor
Homework Helper
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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
2,209
1
[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 thats as good as it gets if I didnt make any errors.
 
Last edited:
  • #11
dextercioby
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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
dextercioby
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[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
dextercioby
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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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