Some techniques one can use to integrate otherwise tough integrals is to create a double integral and use it to replace something in the original integrand.
The LaPlace transform has the form:
\(\displaystyle \int_{0}^{\infty}e^{-st}f(t)dt\)
The thing to do is to transform your integral into someting like this:
I will use t instead of u. It really does not matter, but the Laplace is usually done with a t.
Note that \(\displaystyle \frac{1}{t^{2}}=\int_{0}^{\infty}se^{-st}ds\)
So, we can write:
\(\displaystyle \int_{0}^{\infty}\int_{0}^{\infty}se^{-st}sin^{2}(t)dsdt\)
Switch the order of integration:
\(\displaystyle \int_{0}^{\infty}\int_{0}^{\infty}se^{-st}sin^{2}(t)dtds\)
Integrating by parts we get:
***\(\displaystyle \left\left(\frac{-sin(2t)s}{s^{2}+4}-\frac{sin^{2}(t)s^{2}}{s^{2}+4}-\frac{2}{s^{2}+4}\right)e^{-st}\right|_{0}^{\infty}\)
Using the integration limits, this all disappears except for \(\displaystyle \frac{2}{s^{2}+4}\).
So, we are left with:
\(\displaystyle \int_{0}^{\infty}\frac{2}{s^{2}+4}ds=\left tan^{-1}(\frac{s}{2})\right|_{0}^{\infty}=\frac{\pi}{2}\)
*** we can avoid that nasty integration by parts by considering the imaginary part.
\(\displaystyle s\int_{0}^{\infty}e^{-st}sin^{2}(t)dt=\int_{0}^{\infty}e^{-st}sin(2t)dt=Im\int_{0}^{\infty}e^{-(s-2i)t}dt=Im\frac{1}{s-2i}=\frac{2}{s^{2}+4}\)
If you're wondering why there is a change from \(\displaystyle sin^{2}(t)\) to \(\displaystyle sin(2t)\).
The Laplace of \(\displaystyle sin^{2}(t)\) is \(\displaystyle \frac{2}{s(s^{2}+4)}\)
If this is multiplied by 's', we get the Laplace tranform for sin(2t), which is \(\displaystyle \frac{2}{s^{2}+4}\)
Actually, you could probably skip all this and jump right to the transform, if you wish, by noting the relation between the laplace of \(\displaystyle sin(2t)\) and \(\displaystyle sin^{2}(t)\).
