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Mathematics · Ch 11 — Definite Integration

Reduction Formulae for Definite Integrals of sin^n x and cos^n x

11.2.3

Reduction Formulae for Definite Integrals of sin^n x and cos^n x

For the special family of integrals ∫0π/2sin⁡nx dx\int_0^{\pi/2}\sin^n x\,dx and ∫0π/2cos⁡nx dx\int_0^{\pi/2}\cos^n x\,dx (products of the same power of sine or cosine over a quarter period), there is a standard shortcut called a reduction formula that gives the value directly from nn, without repeated integration by parts:

∫0π/2sin⁡nx dx=(n−1)n⋅(n−3)(n−2)⋅(n−5)(n−4)⋯23,n odd,\int_0^{\pi/2}\sin^n x\,dx=\frac{(n-1)}{n}\cdot\frac{(n-3)}{(n-2)}\cdot\frac{(n-5)}{(n-4)}\cdots\frac{2}{3}, \quad n \text{ odd},

∫0π/2sin⁡nx dx=(n−1)n⋅(n−3)(n−2)⋅(n−5)(n−4)⋯12⋅π2,n even.\int_0^{\pi/2}\sin^n x\,dx=\frac{(n-1)}{n}\cdot\frac{(n-3)}{(n-2)}\cdot\frac{(n-5)}{(n-4)}\cdots\frac{1}{2}\cdot\frac{\pi}{2}, \quad n \text{ even}.

The pattern is: start with n−1n\frac{n-1}{n} and keep multiplying by the next fraction obtained by subtracting 22 from both the numerator's two terms and the denominator's two terms, all the way down to 23\frac23 if nn is odd (no extra factor), or down to 12\frac12 followed by one extra factor of π/2\pi/2 if nn is even.

Reducing cosine to sine. By Property V (Section 4.2.1) with the reflection x→π/2−xx\to \pi/2-x,

∫0π/2cos⁡nx dx=∫0π/2[cos⁡(π2−x)]ndx=∫0π/2sin⁡nx dx,\int_0^{\pi/2}\cos^n x\,dx=\int_0^{\pi/2}\Big[\cos\Big(\frac\pi2-x\Big)\Big]^n dx=\int_0^{\pi/2}\sin^n x\,dx,

so the same reduction formula, with the same value of nn, gives ∫0π/2cos⁡nx dx\int_0^{\pi/2}\cos^n x\,dx as well — there is no need to memorise a second formula.

Worked check for n=7n=7 (odd). ∫0π/2sin⁡7x dx=67⋅45⋅23=6⋅4⋅27⋅5⋅3=1635\displaystyle\int_0^{\pi/2}\sin^7x\,dx=\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}=\frac{6\cdot4\cdot2}{7\cdot5\cdot3}=\frac{16}{35} (the chain stops at 23\frac23 since 77 is odd, with no leftover π/2\pi/2 factor). …

Misc 1Worked value for n = 7 and n = 8

Worked out. The textbook applies the reduction formula to two concrete cases directly after stating it: for the odd power ∫0π/2sin⁡7x dx\int_0^{\pi/2}\sin^7x\,dx the alternating product of consecutive odd/even factors gives 6⋅4⋅27⋅5⋅3=1635\frac{6\cdot4\cdot2}{7\cdot5\cdot3}=\frac{16}{35}, and for the even power ∫0π/2cos⁡8x dx\int_0^{\pi/2}\cos^8x\,dx the same pattern carries an extra factor of π/2\pi/2, giving 7⋅5⋅3⋅18⋅6⋅4⋅2⋅π2=35π256\frac{7\cdot5\cdot3\cdot1}{8\cdot6\cdot4\cdot2}\cdot\frac{\pi}{2}=\frac{35\pi}{256}; these two worked numbers are what a student chec …