We already know we can do the following problem using u$-$substitution, with

$u=sin(-2x)$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^{10}(-2x)cos(-2x)dx=$

$$

We see that ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}si{n}^{10}(-2x)co{s}^3(-2x)dx$ is solved similarly, because

A$.$ the exponent of cos is odd.

B$.$ the exponent of $sin$ is odd.

C$.$ None of the above