Don't know to handle the (ωt) and (t) that appear in the integral

[Karol]

Homework Statement

I have the integral:
[tex]\int_{0}^{4} \sin(\omega t)\cdot t^2[/tex]

Homework Equations

I know from integrals tables that:
[tex]\int x^2 \sin x=2x\sin x-(x^2-2)\cos x[/tex]
But i don't know to handle the (ωt) and (t) that appear in the integral

The Attempt at a Solution

 

[voko]

Homework Statement

I have the integral:
[tex]\int_{0}^{4} \sin(\omega t)\cdot t^2[/tex]

Very simple: $$

\int_{0}^{4} \sin(\omega t)\cdot t^2 d \omega = t^2 \int_0^4 \sin (\omega t) d \omega


= t^2 \left[ \frac {-\cos (\omega t)} {t} \right]_0^4 = t (1 - \cos 4t)
$$

 

[Karol]

Sorry, i didn't mention it's not dω, but dt, the variant is the time, the ω is a constant

 

[Infrared]

Hint: Write [itex] t^2= \frac{1}{\omega^2}(\omega t)^2 [/itex].

 

[voko]

Sorry, i didn't mention it's not dω, but dt, the variant is the time, the ω is a constant

Do you see why it is important to append dx to the integrand? Without this, no one can be sure what variable is being integrated.

 

[Karol]

So, i use [itex] t^2= \frac{1}{\omega^2}(\omega t)^2 [/itex], and then i define a new variable x=ωt and then dx=ωdt, right?

 

[Infrared]

That should work.

 

[HallsofIvy]

Simplest is to use the substitution [itex]y= \omega t[/itex] so that [itex]dy= \omega dt[/itex], [itex]dt= dy/omega[/itex] and [itex]t= y/\omega[/itex]

 

[Karol]

Result

The integral is:
[tex]\int_{0}^{4} \sin(\omega t)\cdot t^2[/tex]
I used:
[tex]t^2= \frac{1}{\omega^2}(\omega t)^2 [/tex]
Then x=ωt and then dx=ωdt.
ω=7.27E-5[rad/sec]
for t=o --> x=0
for t=4 --> x=ωx4[sec]=0.00029
[tex]\frac{1}{\omega^3}=\frac{1}{(7.27E-5)^3}=2.6E12[/tex]
From integrals tables:
[tex]\int x^2 \sin x=2x\sin x-(x^2-2)\cos x[/tex]
So:
[tex]\frac{1}{\omega^3}\int_{0}^{0.00029}\sin x-(x^2-2)\cos x=211093[/tex]
Can anyone check? the result isn't logical, physically