Answer:
[tex]{ \sf{ \int 4x \cos(2 - 3x) dx = {4x}^{2} \cos(2 - 3x) + c }}[/tex]
Step-by-step explanation:
From integration by parts:
[tex]{ \boxed{ \bf{ \int u \: \frac{dv}{dx} = uv - \int v \: \frac{du}{dx} }}}[/tex]
let u be cos(2-3x) and let dv/dx be 4x:
[tex]{ \sf{ \frac{du}{dx} = 3 \sin(2 - 3x) }} \\ \\ { \sf{ v = \int 4x = 2 {x}^{2} }}[/tex]
Substitute in formular box:
[tex]{ \sf{ = (2 {x}^{2} \times \cos(2 - 3x)) - \int 2 {x}^{2} .3 \sin(2 - 3x) dx}} \\ = { \sf{2 {x}^{2} ( \cos(2 - 3x) - 3 \int \sin(2 - 3x)dx }} \\ { \sf{ = 2 {x}^{2} ( \cos(2 - 3x) - 3( - \frac{1}{3} \cos(2 - 3x) ) + c}} \\ { \sf{ = {4x}^{2} \cos(2 - 3x) + c}}[/tex]
