Reduction Formula for Integral of Power of Sine/Mistake

Source Work

 * Appendix $8$: Integrals
 * Reduction Formulae
 * Reduction Formulae

Mistake

 * ''For $I_n = \int \sin^n x \rd x$, where $n \ge 2$, then
 * $I_n = -\dfrac {\sin^n x \cos x} n + \dfrac {n - 1} n I_{n - 2}$.''

Correction
The correct expression is:
 * $I_n = -\dfrac {\sin^{n - 1} x \cos x} n + \dfrac {n - 1} n I_{n - 2}$