For $I_n = \int \sin^n ax\, dx$, where $n$ is a positive integer and $a$ is a constant, which reduction formula is correct?

Differentiating the product gives $\frac{d}{dx}\left(\sin^{n-1}ax\cos ax\right)=(n-1)a\sin^{n-2}ax\cos^2ax-a\sin^n ax$. Replacing $\cos^2ax$ by $1-\sin^2ax$ turns the right side into $(n-1)a\sin^{n-2}ax-na\sin^n ax$. Integrating both sides, $\sin^{n-1}ax\cos ax=(n-1)aI_{n-2}-naI_n$, so $I_n=-\frac{1}{na}\sin^{n-1}ax\cos ax+\frac{n-1}{n}I_{n-2}$. The leading term carries a minus sign, and the $\frac{1}{na}$ shows that the constant $a$ divides only that first term.