Reduction Formula for Integral of Power of Sine/Mistake
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Source Work
2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.):
- Appendix $8$: Integrals
- 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:
- $I_n = -\dfrac {\sin^{n - 1} x \cos x} n + \dfrac {n - 1} n I_{n - 2}$
Sources
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $8$: Integrals: Reduction Formulae