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

1. May 4, 2013

### Karol

1. The problem statement, all variables and given/known data
I have the integral:
$$\int_{0}^{4} \sin(\omega t)\cdot t^2$$

2. Relevant equations
I know from integrals tables that:
$$\int x^2 \sin x=2x\sin x-(x^2-2)\cos x$$
But i don't know to handle the (ωt) and (t) that appear in the integral

3. The attempt at a solution

2. May 4, 2013

### voko

Very simple: $$\int_{0}^{4} \sin(\omega t)\cdot t^2 d \omega = t^2 \int_0^4 \sin (\omega t) d \omega = t^2 \left[ \frac {-\cos (\omega t)} {t} \right]_0^4 = t (1 - \cos 4t)$$

3. May 4, 2013

### Karol

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

4. May 4, 2013

### Infrared

Hint: Write $t^2= \frac{1}{\omega^2}(\omega t)^2$.

5. May 4, 2013

### voko

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.

6. May 4, 2013

### Karol

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

7. May 4, 2013

### Infrared

That should work.

8. May 4, 2013

### HallsofIvy

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

Last edited by a moderator: May 4, 2013
9. May 5, 2013

### Karol

Result

The integral is:
$$\int_{0}^{4} \sin(\omega t)\cdot t^2$$
I used:
$$t^2= \frac{1}{\omega^2}(\omega t)^2$$
Then x=ωt and then dx=ωdt.
$$\frac{1}{\omega^3}=\frac{1}{(7.27E-5)^3}=2.6E12$$
$$\int x^2 \sin x=2x\sin x-(x^2-2)\cos x$$
$$\frac{1}{\omega^3}\int_{0}^{0.00029}\sin x-(x^2-2)\cos x=211093$$