How do you use the second fundamental theorem of Calculus to find the derivative of given int sin^3(2t-1) dt from [x-1, x]?

Hence we need to find the derivative of

${\int}_{x - 1}^{x} {\sin}^{3} \left(2 t - 1\right) \mathrm{dt}$

So we use the Leibniz Integral Rule which in our case becomes

$\frac{d}{\mathrm{dx}} \left[{\int}_{a \left(x\right)}^{b \left(x\right)} f \left(t\right) \mathrm{dt}\right] = f \left(b \left(x\right)\right) \cdot \frac{\mathrm{db} \left(x\right)}{\mathrm{dx}} - f \left(a \left(x\right)\right) \cdot \frac{d \left(a \left(x\right)\right)}{\mathrm{dx}}$

Hence the derivative for $b \left(x\right) = x$ ,$a \left(x\right) = x - 1$ ,$f \left(t\right) = {\sin}^{3} \left(2 t - 1\right)$

is

$\frac{d}{\mathrm{dx}} \left({\int}_{x - 1}^{x} {\sin}^{3} \left(2 t - 1\right) \mathrm{dt}\right) = {\sin}^{3} \left(2 x - 1\right) \cdot \left(x\right) ' - {\sin}^{3} \left(2 \left(x - 1\right) - 1\right) \cdot \left(x - 1\right) ' = {\sin}^{3} \left(2 x - 1\right) - {\sin}^{3} \left(2 x - 3\right)$