Question:
Let $f=e^{-kt}\sin^2(\omega t)$. Find $\langle f\rangle$ for one cycle from $t=0$ to $2\pi/\omega$.
My attempt:
I wished to do by first principle,
$$\langle f\rangle=\frac{\int_0^{2\pi/\omega}f d t}{\int_0^{2\pi/\omega}d t}$$
However, this involves use of integration by parts method and thus is highly convoluted. The answer in fact, is simply:
$$\langle f\rangle=\frac{e^{-kt}}{2}$$
which suggests to me that there is an alternate and more intuitive method to arrive at the result. I wished to know this method.
PS: The answer expression I've stated above has been given to me by my professor as is. I tried to contact him regarding the exact value of the $t$ in that expression, but he was unavailable for any further comment. The more experienced mathematicians among you may immediately spot a typo in the result if there's any, and hence please correct the same. Thank you!