Integrals l

$\begin{aligned} I & = \int 2(3x-2)\sin^5 (x^3-2x) \cos (x^3-2x) \ dx & \small \color{#3D99F6} \text{Let }u = \sin (x^3-2x) \\ & = \int 2u^5 \ du & \small \color{#3D99F6} \implies du = (3x-2)\cos (x^3-2x) \ dx \\ & = \frac {u^6}3 + C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = \boxed{\frac {\sin^6 (x^3-2x)}3+C} \end{aligned}$

$\int 2(3x^2 - 2) \, sin^5 (x^3 - 2x) \, cos(x^3 - 2x) \, dx$

Variable substitution:

$u = x^3 - 2x \\ dx = \frac{du}{3x^2 - 2}$

Plugging in:

$\int 2 \, sin^5 (u) \, cos (u) \, du$

Another variable substitution:

$v = sin (u )\\ du = \frac{dv}{cos(u)}$

Plugging in:

$\int 2 \, v^5 \, dv = \frac{v^6}{3} + C = \frac{sin^6 (u)}{3} + C = \frac{sin^6 (x^3 - 2x)}{3} + C$