Calculating the integral periodic functions times the heaviside function

Question

$$\int_0^{t'}sin \left(2t-\frac{\pi}{4} \right)H(t)dt'$$

I tried integrating by parts:

$$dg = sin\left(2t-\frac{\pi}{4} \right) \rightarrow g = -\frac12cos\left(2t-\frac{\pi}{4} \right) $$

$$ f = H(t) \rightarrow df = \delta(t)$$

Plugging everything in:

$$ -H(t)\left(\frac12cos\left(2t-\frac{\pi}{4} \right)\right)|^{t'}_{0}+\int_0^{t'} \delta(t)\frac12cos\left(2t-\frac{\pi}{4} \right) dt'$$

What is the right most integral give me? It seems as though I can't apply any of the properties of the delta function to this integral

Answer 1

The Heaviside function is equal to 1 in the entire interior of the integral's domain (assuming $t'>0$). So $$\int_0^{t'}sin \left(2t-\frac{\pi}{4} \right)H(t)dt'=\int_0^{t'}sin \left(2t-\frac{\pi}{4} \right)dt'$$