Why can't I solve $\int \sin^3(x)dx$ directly using u-substitution without trig identities?

Question

I saw that the standard solution to the problem

$\int \sin^3(x)dx$

involves first using trig identities and then u-substitution. I am not understanding why it is wrong to use directly u-substitution, such as:

$u=\sin x$

$du=\cos x dx$

$\frac{1}{\cos x}\int \sin^3(x)\cdot \cos x dx =\frac{1}{\cos x} \int u^3du= \frac{\sin^4(x)}{4\cos x}$