Definite integrals and reduction formulae

Some integrals refuse to sit still for a single antiderivative. Ask for $\int \sin^7 x\,dx$ or $\int x^9 e^x\,dx$ and repeating integration by parts nine separate times is not a plan, it is a punishment. Reduction formulae turn that punishment into a one-line recursion: express the integral $I_n$ at power $n$ in terms of the same integral at a lower power, $I_{n-2}$ or $I_{n-1}$ , and then unwind the chain arithmetically instead of integrating again and again. This node builds that machine for powers of sine, and the closely related family for tangent, and shows the payoff at $[0,\pi/2]$ : the boundary terms vanish and the recursion collapses into a clean closed form called Wallis's formula. You already have integration by parts and the Pythagorean identity — everything here is one clever substitution, applied to the same pattern, again and again.

The core insight

A reduction formula writes $I_n$ in terms of $I_{n-2}$ (or $I_{n-1}$ ) via one integration-by-parts step, so a high-power integral collapses to a short arithmetic recursion instead of repeated integration.

The core idea

What a reduction formula is. For a family $I_n = \int f_n(x)\,dx$ , integration by parts on $f_n = f_{n-1}\cdot f$ typically reproduces $I_n$ itself on the right-hand side; solving that equation algebraically for $I_n$ leaves it expressed through $I_{n-2}$ (or $I_{n-1}$ ) plus a boundary/algebraic term. This is a technique applied once, not a table of formulae to memorize separately for every function. The payoff is computational: instead of doing $n$ separate integration-by-parts calculations to reach $\int \sin^{9}x\,dx$ , you derive the recursion once and then apply arithmetic — multiply by a fraction, drop the index by two — four or five times in a row.

The sine family. Splitting $\sin^n x = \sin^{n-1}x\cdot\sin x$ and integrating by parts gives

I_n = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}I_{n-2},

where: $I_n = \int \sin^n x\,dx$ , and the boundary term $-\sin^{n-1}x\cos x/n$ carries all the $x$ -dependence left over after one integration by parts.

The clean definite case. Over $[0,\pi/2]$ the boundary term is zero at both ends (since $\sin 0=0$ and $\cos(\pi/2)=0$ ), so

\int_0^{\pi/2}\sin^n x\,dx = \frac{n-1}{n}\int_0^{\pi/2}\sin^{n-2}x\,dx,

and because $\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\cos^n x\,dx$ (substitute $x \to \pi/2 - x$ ), unwinding this recursion all the way down gives Wallis's formula:

\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\cos^n x\,dx = \begin{cases} \dfrac{(n-1)(n-3)\cdots 2}{n(n-2)\cdots 3}, & n \text{ odd} \\[2mm] \dfrac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}\cdot \dfrac{\pi}{2}, & n \text{ even.} \end{cases}

The tangent family. Using $\tan^n x = \tan^{n-2}x(\sec^2 x - 1)$ instead of the Pythagorean substitution for $\sin^2$ , integration gives a simpler recursion with no boundary term at all:

I_n = \frac{\tan^{n-1}x}{n-1} - I_{n-2}.

The figure below shows the recursion as a chain: each arrow is one integration-by-parts step, and the chain terminates at $I_0$ or $I_1$ , which are elementary.

Figure

A chain diagram showing In linked by arrows to I sub n minus two down to I zero or I one, each arrow labelled one integration by parts step

Derivation

Derivation ADeriving the sine reduction formula by integration by parts

The goal is to show, starting from nothing but integration by parts and the identity $\sin^2 x + \cos^2 x = 1$ , that $I_n = \int \sin^n x\,dx$ satisfies $I_n = -\sin^{n-1}x\cos x/n + \frac{n-1}{n}I_{n-2}$ , so that any power of sine can be integrated by a short recursive computation instead of a fresh integration-by-parts calculation every time.

Step 1. Split the integrand as a product so integration by parts has something to differentiate and something to integrate. Write $\sin^n x = \sin^{n-1}x\cdot \sin x$ and set $u = \sin^{n-1}x$ , $dv = \sin x\,dx$ , so $du = (n-1)\sin^{n-2}x\cos x\,dx$ and $v = -\cos x$ .

Step 2. Apply integration by parts, $\int u\,dv = uv - \int v\,du$ :

I_n = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\cos^2 x\,dx.

Step 3. The remaining integral still has a $\cos^2 x$ in it, which is not yet in terms of $I_{n-2}$ or $I_n$ . Substitute $\cos^2 x = 1 - \sin^2 x$ and split the integral into two pieces:

I_n = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\,dx - (n-1)\int \sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)I_{n-2} - (n-1)I_n.

Step 4. The unknown $I_n$ now appears on both sides of the equation, so collect it algebraically: adding $(n-1)I_n$ to both sides gives $nI_n = -\sin^{n-1}x\cos x + (n-1)I_{n-2}$ . Dividing by $n$ produces the reduction formula

I_n = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}I_{n-2}.

This is the same result quoted in the core idea section above, and the identical algebraic trick — split the product, integrate by parts, substitute the Pythagorean identity, collect $I_n$ — reproduces the reduction formulae for $\cos^n x$ and, with $\sec^2 x - 1$ in place of the Pythagorean substitution, for $\tan^n x$ .

Worked example

Problem. Evaluate $\displaystyle\int_0^{\pi/2}\sin^6 x\,dx$ using Wallis's formula, and cross-check by evaluating $\displaystyle\int_0^{\pi/4}\tan^4 x\,dx$ using the tangent reduction formula.

Step 1. For $\int_0^{\pi/2}\sin^6 x\,dx$ , $n = 6$ is even, so Wallis's formula gives $\dfrac{(6-1)(6-3)(6-5)}{6(6-2)(6-4)}\cdot\dfrac{\pi}{2} = \dfrac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\cdot\dfrac{\pi}{2} = \dfrac{15}{48}\cdot\dfrac{\pi}{2} = \dfrac{5\pi}{32}$ .

Step 2. Check this by direct recursion: $\int_0^{\pi/2}\sin^6 x\,dx = \frac{5}{6}\int_0^{\pi/2}\sin^4 x\,dx = \frac{5}{6}\cdot\frac{3}{4}\int_0^{\pi/2}\sin^2 x\,dx = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{15}{48}\cdot\frac{\pi}{2} = \frac{5\pi}{32}$ . Both routes agree.

Step 3. For $\int \tan^n x\,dx$ , apply $I_n = \frac{\tan^{n-1}x}{n-1} - I_{n-2}$ twice: $I_4 = \frac{\tan^3 x}{3} - I_2$ , and $I_2 = \tan x - I_0 = \tan x - x$ (since $I_0 = \int 1\,dx = x$ ). So $I_4 = \frac{\tan^3 x}{3} - \tan x + x$ .

Step 4. Evaluate between $0$ and $\pi/4$ , using $\tan(\pi/4) = 1$ : $\int_0^{\pi/4}\tan^4 x\,dx = \left[\frac{\tan^3 x}{3} - \tan x + x\right]_0^{\pi/4} = \left(\frac13 - 1 + \frac{\pi}{4}\right) - 0 = \frac{\pi}{4} - \frac{2}{3} \approx 0.7854 - 0.6667 = 0.1187$ .

Final answers: $\displaystyle\int_0^{\pi/2}\sin^6 x\,dx = \frac{5\pi}{32}$ , and $\displaystyle\int_0^{\pi/4}\tan^4 x\,dx = \frac{\pi}{4} - \frac{2}{3} \approx 0.1187$ .

Competitive stretch

For odd $n$ , Wallis's formula never carries a $\pi/2$ factor — try $\int_0^{\pi/2}\sin^5 x\,dx = \frac{4\cdot 2}{5\cdot 3}\cdot 1 = \frac{8}{15}$ , and notice how different the odd- $n$ pattern looks from the even- $n$ one above; mixing the two cases up is the single most common Wallis mistake.

Common traps

Applying the clean Wallis formula (with no boundary term) to an indefinite integral — the boundary term $-\sin^{n-1}x\cos x/n$ only vanishes for the definite integral over $[0,\pi/2]$ .
Swapping the odd- $n$ and even- $n$ cases of Wallis's formula: odd $n$ ends the product at $1$ with no $\pi/2$ factor, even $n$ ends at $1$ and does carry a $\pi/2$ factor.
Getting the sign wrong in the tangent recursion — it is $I_n = \tan^{n-1}x/(n-1) - I_{n-2}$ , a minus sign, not a plus.
Reducing all the way down to $I_0$ or $I_1$ by hand for large $n$ instead of just plugging into the closed-form Wallis product, which is faster and less error-prone for a single numeric answer.
Forgetting that $\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\cos^n x\,dx$ only holds over this specific symmetric interval — the substitution $x \to \pi/2 - x$ that proves it does not carry over to an arbitrary interval like $[0,\pi]$ .

Ask the tutor: if collecting the $I_n$ terms algebraically in Step 4 of the derivation feels confusing, ask the AI tutor to redo that step slowly with the cosine or tangent version side by side.

Why does the boundary term vanish specifically on 0 to pi over 2 and not on some other interval?

The boundary term is $-\sin^{n-1}x\cos x/n$ , evaluated at the two limits. At $x=0$ , $\sin 0 = 0$ kills the term outright (for $n\ge 1$ ). At $x=\pi/2$ , $\cos(\pi/2)=0$ kills it there too. Any interval where both endpoints make either $\sin$ or $\cos$ vanish would work in principle, but $[0,\pi/2]$ is the natural, exam-standard choice because it is also where the sine and cosine reduction formulae coincide by symmetry.

Do I need to memorize the sine, cosine and tangent reduction formulae separately?

Not really — memorize the method: split the highest power off, integrate by parts (or, for tangent, use $\sec^2 x - 1$ ), then substitute the Pythagorean identity and collect terms. If you can reproduce the sine derivation from scratch in under two minutes, you can rebuild any of the others the same way under exam pressure, which is safer than trying to recall three separate formulae verbatim.

Is there a reduction formula for mixed products like the integral of sin to the m x cos to the n x?

Yes, and it follows the exact same template: split off one factor of $\sin x$ or $\cos x$ , integrate by parts, and use the Pythagorean identity to reduce either $m$ or $n$ by two while the other index stays fixed. It is not covered in worked form here since it is a direct extension of the sine and tangent cases above, but the underlying technique — split, integrate by parts, substitute, collect — is identical.