Making sense of Absolute value in an integral

Question

In the following integral: $$\int_0^\infty |e^{-10t}\cos(30\pi t)|^2dt$$ May I say that $|e^{-10t}\cos(30\pi t)|=|e^{-10t}||\cos(30\pi t)|=e^{-10t}\cos(30\pi t)$?

If not how can I expand upon $|e^{-10t}\cos(30\pi t)|^2$?

Answer 1

HINT

Use:

$$\left|e^{-10t}\cos(30\pi t)\right|^2=e^{-20\Re(t)}\left|\cos(30\pi t)\right|^2$$

Now, when $t\in\mathbb{R}$:

$$e^{-20\Re(t)}\left|\cos(30\pi t)\right|^2=e^{-20t}\cos^2(30\pi t)$$

So, we get (using a substitution $u=10t$ and $\text{d}u=10\space\text{d}t$):

$$\text{I}=\int_0^\infty\left|e^{-10t}\cos(30\pi t)\right|^2\space\text{d}t=\int _0^\infty e^{-20t}\cos^2(30\pi t)\space\text{d}t=\frac{1}{10}\int_0^\infty e^{-2u}\cos^2(3\pi u)\space\text{d}u$$

Now, use (and apply linearity):

$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$

To get:

$$\text{I}=\frac{1}{20}\int_0^\infty e^{-2u}\left(1+\cos(6\pi u)\right)\space\text{d}u$$

Use again a substitution $s=2u$ and $\text{d}s=2\text{d}u$:

$$\text{I}=\frac{1}{40}\int_0^\infty e^{-s}\left(1+\cos(3\pi s)\right)\space\text{d}s$$