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

In summary, you need to use substitution to turn the integral into one with respect to u instead of t. The conversion factor is right above the integral.
  • #1
PhysicsMajor
15
0
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:
 
Physics news on Phys.org
  • #2
[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
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
From converting the integral to one with respect to u instead of t. The conversion factor is right above the integral
 
  • #5
Ugh I did it wrong hold on. I am so tired
 
  • #6
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
Mathworld's Integrator gives a long and not-very-nice answer btw.
 
  • #8
Then you'd have to raise that to the fifth and foil and blah blah blah. Its a real complicated integral.
 
  • #9
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.
 

What is the meaning of the integral of (sin 3t)^5 cos t dt?

The integral of (sin 3t)^5 cos t dt represents the area under the curve of the function (sin 3t)^5 cos t, with respect to the variable t. In other words, it is a way to find the total amount of the function's output over a certain interval.

Why is the function (sin 3t)^5 cos t used in this integral?

The function (sin 3t)^5 cos t is used because it is a common example of an integrable function. It also has a relatively simple form, making it easier to calculate the integral.

What is the process for finding the integral of (sin 3t)^5 cos t dt?

The process for finding the integral of (sin 3t)^5 cos t dt involves using integration techniques, such as u-substitution and trigonometric identities, to simplify the function and then applying the fundamental theorem of calculus to solve the integral.

Is it necessary to use calculus to solve this integral?

Yes, it is necessary to use calculus to solve this integral. Integration is a fundamental concept in calculus, and this integral cannot be solved without using calculus principles.

What are some real-world applications of this integral?

The integral of (sin 3t)^5 cos t dt has various real-world applications, such as in physics and engineering to calculate the total energy or work done by a varying force over a certain interval. It can also be used in finance to calculate compound interest or in economics to calculate total revenue or profit.

Similar threads

  • Calculus
Replies
3
Views
2K
Replies
8
Views
425
Replies
16
Views
1K
Replies
8
Views
176
  • Calculus
Replies
6
Views
1K
Replies
3
Views
1K
  • Calculus
Replies
29
Views
718
Replies
4
Views
1K
Replies
2
Views
291
Back
Top