Change bounds of integral

Question

I want to change the bounds of this integral from $0,2t$ to $0,t$ (for Laplace transform)

I try using an extra variable $x=2t$ but I can't realize how to change the $\sin:$

$$ f(t) = \int_0^{2t} \sin (u)\, du. $$

Answer 1

Now $u \in (0,2t)$ and let $v = u/2 \in (0,t)$ so you have $$ \int_0^{2t} \sin(u) du = \int_0^t \sin(2v) d(2v) = 2\int_0^t \sin(2v)dv. $$