Solve F ( x ) = ∫ (from 0 to x ^ 3) of -3x^2f(t)dtquadtext{e}f(t)=int_{0}^1+frac{pi{2}}sin^3(u)*cos(u)du

$$F ( x ) = \int _ { 0 } ^ { x ^ { 3 } - 3 x ^ { 2 } } f ( t ) d t \quad \text { e } f ( t ) = \int _ { 0 } ^ { 1 + \frac { \pi } { 2 } } \sin ^ { 3 } ( u ) \cdot \cos ( u ) d u$$

Evaluate (complex solution)

$$Fx=\frac{etx^{4}\left(f\left(x-3\right)\right)^{2}}{2}\text{ and }\frac{etx^{4}\left(f\left(x-3\right)\right)^{2}}{2}=\frac{\left(\cos(2)+1\right)^{2}}{16}$$

Solve for t

$$t=\frac{\left(\frac{\cos(2)+1}{f\left(x-3\right)}\right)^{2}}{8ex^{4}}$$

$$x\neq 3\text{ and }f\neq 0\text{ and }F=\frac{\left(\cos(2)+1\right)^{2}}{16x}\text{ and }x\neq 0$$