Integrating the product of a step function and a trigonometric function

[Rome_Leader]

Homework Statement

Integrate _-π∫^π H(t-π/2)*sin(2t)dt

Homework Equations

See above.

The Attempt at a Solution

I can rationalize the slightly simpler integral for the same limits of H(t)*sin(2t) as coming out to 0 due to the definition of the unit step function, but I'm wondering if the subtraction of π/2 changes it any. It's still integrating over the range of H(t), correct? So should it still work out to be 0?

 

[HallsofIvy]

$H(t- \pi/2)$ is equal to 0 for $t< \pi/2$, 1 for $t\ge \pi/2$. So that integral is just
$$\int_{\pi/2}^\pi sin(2t)dt$$