I can't figure out how you're supposed to find the solutions for a product involving trigonometric functions with heaviside, most examples online involve exponentials which have the nice property of being their own derivative/integral.
$$\int \sin(3t)\theta(t)dt $$
Wolfram gives me the following answer.
I know you can split up definite integrals piecewise where heaviside evaluates to 0 and 1 respectively, but how would you go about solving indefinite integrals in this way, especially trigonometric ones?
Best Answer
Integrate by parts,
$$\begin{align} &\int \sin3t\theta(t)dt =-\frac13\int d(\cos3t)\theta(t)dt \\ =&-\frac13\theta(t)\cos3t+\frac13\int \cos3t\delta(t)dt\\ =&-\frac13\theta(t)\cos3t+\frac13\theta(t)\cos(0)\\ =&\frac13\theta(t)(1-\cos3t)=\frac23\theta(t)\sin^2\frac{3t}2 \end{align}$$
where $\delta(t)$ is the Dirac delta function, the derivative of the Heaviside function $\theta(t)$.