d/dx $\displaystyle \int_x^0 \! sin^2(t) \, \mathrm{d}x.$

so first I flipped it & changed the sign to get the x on top:

d/dx - $\displaystyle \int_0^x \! sin^2(t) \, \mathrm{d}x.$

Now my question is, what is the integral of sin^2(t)?

Like I know for just integral sinxdx=-cosx...

If I set cos2x=1-sin^2(x)..will this work?