From a Reduction Formula to a Definite Integral
Reduction formulas replace an integral with a related integral of lower complexity. For sine powers, the exponent drops by two at each step. On a definite interval, the endpoint values matter: they determine whether integration by parts leaves an additional term. On \([0,\pi/2]\), the endpoint term vanishes for the exponents used in the recurrence, giving a particularly simple calculation.
We will use that recurrence to evaluate every integral of a nonnegative integer power of sine on this interval. Repeated reduction ends at one of two base cases, depending on whether the original exponent is even or odd. Those different base cases lead to two distinct finite-product formulas.
Define the Integral and Its Base Cases
For each integer \(n\geq 0\), define \(I_n\) to be the definite integral of \(\sin^n x\) from \(0\) to \(\pi/2\). The integrand is continuous on this closed interval, so the integral exists.
The first two values can be calculated directly. Since \(\sin^0 x=1\), the exponent-zero integral is the length of the interval. For exponent one, an antiderivative of \(\sin x\) is \(-\cos x\).
These base cases are not interchangeable. Reducing an exponent by two preserves whether it is even or odd, so an even exponent eventually reaches \(I_0\), while an odd exponent eventually reaches \(I_1\).
The Definite Sine-Power Recurrence
To derive the recurrence, integrate by parts with \(u=\sin^{n-1}x\) and \(dv=\sin x\,dx\). Then \(du=(n-1)\sin^{n-2}x\cos x\,dx\) and \(v=-\cos x\). The boundary term is zero: at \(x=\pi/2\), \(\cos x=0\); at \(x=0\), \(\sin^{n-1}x=0\) because \(n\geq2\).
The last equality uses \(\cos^2x=1-\sin^2x\), then separates the resulting integral into \(I_{n-2}\) and \(I_n\). Rearranging gives \(nI_n=(n-1)I_{n-2}\), and division by \(n\), which is nonzero for \(n\geq2\), proves the stated recurrence.
This proof also shows why the restriction \(n\geq2\) is important. If \(n=1\), the boundary expression at \(x=0\) is \(-\sin^0(0)\cos(0)=-1\), not zero. The recurrence does not apply at \(n=1\); \(I_1\) is instead one of the base cases.
Finite Products for Even and Odd Exponents
For an even exponent, apply the recurrence repeatedly until the exponent reaches zero. If \(n=2m\), where \(m\) is a nonnegative integer, each step contributes a factor, and the final integral is \(I_0\). Thus the product contains the odd integers from \(1\) through \(2m-1\) in its numerator and the even integers from \(2\) through \(2m\) in its denominator.
For \(m=0\), the product has no factors and is interpreted as \(1\), giving the correct base value \(I_0=\pi/2\). For an odd exponent \(n=2m+1\), repeated reduction ends instead at \(I_1=1\). The resulting product has even factors in the numerator and odd factors in the denominator.
When \(m=0\), the numerator is an empty product, so its value is \(1\). The formula then gives \(I_1=1\). These expressions come directly from repeatedly applying the recurrence; they are finite products, not infinite series or approximations.
Worked Examples
Worked Example: Evaluate \(I_6\)
Since \(6\) is even, reduce successively to \(I_0\). At each step, the current exponent \(n\) contributes the factor \((n-1)/n\).
Check the result by applying the recurrence from \(I_4\). The same reduction gives \(I_4=(3/4)(1/2)I_0=3\pi/16\). Therefore \((5/6)I_4=(5/6)(3\pi/16)=5\pi/32\), agreeing with the calculation of \(I_6\).
Worked Example: Evaluate \(I_7\)
Since \(7\) is odd, the recurrence ends at \(I_1=1\). Reduce two powers at a time and keep the factor corresponding to each exponent.
For a check, first compute \(I_5=(4/5)(2/3)I_1=8/15\). Then the recurrence gives \(I_7=(6/7)I_5=(6/7)(8/15)=48/105=16/35\), as above. The final value is dimensionless because the variable and interval here are angular quantities in radians.
Worked Example: Evaluate \(I_8\) from the Finite Product
The even-exponent product for \(I_8=I_{2\cdot4}\) has four factors. Writing each factor explicitly prevents a missed or misplaced term.
Check with the recurrence from \(I_6\). The previous example gives \(I_6=5\pi/32\), so \((7/8)I_6=(7/8)(5\pi/32)=35\pi/256\), which matches. This also checks that the last reduction factor is \(7/8\), not \(8/7\).
Why the Two Products Differ
The parity distinction follows from the step size in the recurrence. Starting at an even exponent produces the sequence \(2m,2m-2,\ldots,2,0\). Starting at an odd exponent produces \(2m+1,2m-1,\ldots,3,1\). In both cases, each recurrence factor is formed from the current exponent, but the endpoint base value differs: \(I_0=\pi/2\) contributes a factor of \(\pi/2\) to every even-power answer, whereas \(I_1=1\) contributes no extra factor to the odd-power answer.
This distinction is useful as a quick consistency check. An even-power formula should reduce to \(\pi/2\) when \(m=0\); an odd-power formula should reduce to \(1\). If a proposed product fails either test, its factors or base case have been mishandled.
Common Mistake: Losing the Sign in the Rearrangement
A frequent error is to stop at \(I_n=(n-1)(I_{n-2}-I_n)\) and solve incorrectly, perhaps writing \(I_n=(n-1)/(n-2)\,I_{n-2}\). The term being moved to the left is \(-(n-1)I_n\), so it adds to \(I_n\), not to its coefficient with a minus sign.
The incorrect factor would already fail for \(n=2\): it would require division by \(0\), even though \(I_2\) is finite. The correct recurrence gives \(I_2=(1/2)I_0=\pi/4\). Another related mistake is to use \(I_0\) as the endpoint for an odd exponent. Since the exponent decreases by two, odd exponents terminate at \(I_1\), not \(I_0\).
Using the Recurrence Reliably
When evaluating a particular integral, keep the exponent and its factor visible at every step. For \(I_n\), the first factor is always \((n-1)/n\); after that the exponent becomes \(n-2\), so the next factor is \((n-3)/(n-2)\). Continue until reaching the appropriate base case. This makes the finite product and the final multiplication straightforward to audit.
Decide whether reduction will end at \(I_0\) or \(I_1\).
For each current exponent \(n\geq2\), multiply by \((n-1)/n\) and lower the exponent by two.
Use \(I_0=\pi/2\) for an even exponent and \(I_1=1\) for an odd exponent.
Reduce the product, then verify the result using one recurrence step from the preceding integral.
The recurrence is especially valuable when the exponent is large: it avoids integrating a high power directly and converts the calculation into exact rational factors and one base value. Its reliability depends on two details established above: the recurrence is valid for integer \(n\geq2\), and the endpoint base case must match the parity of \(n\).
Check Your Understanding
Use the recurrence and the base cases to answer the following questions.
- What are the values of \(I_0\) and \(I_1\), and why are both needed?
- State the recurrence for \(I_n\), including the condition on \(n\).
- Write \(I_{2m}\) as a finite product, including its base-value factor.
- Write \(I_{2m+1}\) as a finite product and identify the value reached after repeated reduction.
- Explain why integration by parts does not give the stated recurrence when \(n=1\).
