Your weak spots

Insights load after your first practice round.

Module 8 · L17 of 20 ~40 min ⚡ +95 XP available

Reduction Formulas

When an integral like $\int \sin^n x\,dx$ looks impossible to tackle directly, a reduction formula steps in — it expresses $I_n$ in terms of $I_{n-2}$, transforming a hard integral into a repeating pattern you can unwind step by step. In this lesson you'll derive the core formulas using integration by parts, then apply them to evaluate definite integrals with large powers.

Today's hook — Without computing it, do you think $\int_0^{\pi/2} \sin^6 x\,dx$ is larger or smaller than $\dfrac{1}{2}$? Jot your gut answer — you'll check it after card 07.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

You want to evaluate $\int_0^{\pi/2} \sin^6 x\,dx$. Without looking up a formula — do you think this integral is larger or smaller than $\dfrac{1}{2}$? What strategy would you try first?

auto-saved

The big idea: reduction formulas

+5 XP to read

A reduction formula expresses $I_n = \int f^n(x)\,dx$ in terms of $I_{n-2}$ (or $I_{n-1}$), reducing the power by 2 (or 1) at each step. The process repeats until the integral is simple enough to evaluate directly.

The strategy always involves integration by parts. Write $\sin^n x = \sin^{n-1} x \cdot \sin x$, set $u = \sin^{n-1} x$ and $dv = \sin x\,dx$, differentiate and integrate, then use the Pythagorean identity to rewrite $\cos^2 x = 1 - \sin^2 x$.

Result: $I_n = -\dfrac{\sin^{n-1}x\cos x}{n} + \dfrac{n-1}{n}\,I_{n-2}$

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

For definite integrals on $[0, \pi/2]$

The boundary term $[-\frac{1}{n}\sin^{n-1}x\cos x]_0^{\pi/2}$ always equals 0, leaving the clean form $I_n = \dfrac{n-1}{n}\,I_{n-2}$.

Terminal cases

$I_0 = \int_0^{\pi/2}1\,dx = \dfrac{\pi}{2}$ and $I_1 = \int_0^{\pi/2}\sin x\,dx = 1$. Even $n$ terminates at $I_0$; odd $n$ at $I_1$.

Same formula for cosine

$\int_0^{\pi/2}\cos^n x\,dx = \dfrac{n-1}{n}\,I_{n-2}$ on $[0,\pi/2]$ — identical result by symmetry (substitute $x = \pi/2 - t$).

What you'll master

Know

Key facts

$I_n = -\dfrac{\sin^{n-1}x\cos x}{n} + \dfrac{n-1}{n}\,I_{n-2}$ (indefinite form)
On $[0,\pi/2]$: $I_n = \dfrac{n-1}{n}\,I_{n-2}$, with $I_0 = \dfrac{\pi}{2}$, $I_1 = 1$
The same reduction formula applies to powers of $\cos x$

Understand

Concepts

Why integration by parts with the Pythagorean identity produces the reduction
How the boundary term vanishes on $[0, \pi/2]$
The pattern of even/odd chains terminating at $I_0$ or $I_1$

Can do

Skills

Derive the sine reduction formula from first principles
Evaluate $\int_0^{\pi/2} \sin^n x\,dx$ for small positive integer $n$
Prove given reduction formulas using integration by parts

Key terms

Reduction formulaA recurrence relation that expresses $I_n$ in terms of $I_{n-k}$ for some positive integer $k$, reducing the complexity of the integral.

$I_n$Standard notation for $\int_0^{\pi/2} \sin^n x\,dx$ (or the indefinite version). The subscript tracks the power.

Integration by parts$\int u\,dv = uv - \int v\,du$. The tool used to derive all standard reduction formulas.

Pythagorean identity$\sin^2 x + \cos^2 x = 1$, used to rewrite $\cos^2 x = 1 - \sin^2 x$ when simplifying the IBP result.

Boundary termThe $[uv]_a^b$ part of IBP. On $[0,\pi/2]$ for sine powers this always evaluates to 0, simplifying the reduction.

Terminal caseThe base integral that stops the reduction chain: $I_0 = \pi/2$ for even powers; $I_1 = 1$ for odd powers.

Deriving the sine reduction formula

core concept

Let $I_n = \int \sin^n x\,dx$ where $n \geq 2$. Write $\sin^n x = \sin^{n-1}x \cdot \sin x$ and apply integration by parts:

$$u = \sin^{n-1}x,\quad dv = \sin x\,dx$$

$$du = (n-1)\sin^{n-2}x\cos x\,dx,\quad v = -\cos x$$

Applying $\int u\,dv = uv - \int v\,du$:

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

Substitute $\cos^2 x = 1 - \sin^2 x$:

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

$$I_n = -\sin^{n-1}x\cos x + (n-1)I_{n-2} - (n-1)I_n$$

Collect the $I_n$ terms on the left:

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

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

Definite integral on $[0, \pi/2]$. The boundary term: at $x = \pi/2$, $\cos(\pi/2) = 0$; at $x = 0$, $\sin 0 = 0$. So $\left[-\dfrac{\sin^{n-1}x\cos x}{n}\right]_0^{\pi/2} = 0$, giving the clean recurrence $I_n = \dfrac{n-1}{n}\,I_{n-2}$.

Let $I_n = \int \sin^n x\,dx$ where $n \geq 2$. Write $\sin^n x = \sin^{n-1}x \cdot \sin x$ and apply integration by parts:

Pause — copy the sine reduction formula: $I_n=-\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}I_{n-2}$ and the two-step derivation (split, integrate by parts) into your book.

Quick check: When deriving the sine reduction formula using IBP with $u = \sin^{n-1}x$ and $dv = \sin x\,dx$, what is $v$?

Applying the reduction formula: definite integrals

core concept

We just saw the sine reduction formula derived by parts: $I_n=\int\sin^n x\,dx=-\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}I_{n-2}$. That raises a question: on $[0,\pi/2]$, the boundary terms vanish and the formula telescopes — what are the final closed forms for $\int_0^{\pi/2}\sin^n x\,dx$ for even and odd $n$? This card answers it → even: $I_n=\frac{(n-1)!!}{n!!}\cdot\frac{\pi}{2}$; odd: $I_n=\frac{(n-1)!!}{n!!}$.

On $[0, \pi/2]$, the reduction $I_n = \dfrac{n-1}{n}\,I_{n-2}$ unwinds like a telescope. For even $n$, chain down to $I_0 = \dfrac{\pi}{2}$; for odd $n$, chain down to $I_1 = 1$.

Even example — $I_4$:

$$I_4 = \frac{3}{4}\,I_2 = \frac{3}{4}\cdot\frac{1}{2}\,I_0 = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3\pi}{16}$$

Odd example — $I_5$:

$$I_5 = \frac{4}{5}\,I_3 = \frac{4}{5}\cdot\frac{2}{3}\,I_1 = \frac{4}{5}\cdot\frac{2}{3}\cdot 1 = \frac{8}{15}$$

Notice the pattern: for even $n$, multiply descending odd/even fractions and end with $\dfrac{\pi}{2}$. For odd $n$, multiply and end with $1$.

Wallis's product. The general results $\int_0^{\pi/2}\sin^{2m}x\,dx = \dfrac{(2m-1)!!}{(2m)!!}\cdot\dfrac{\pi}{2}$ and $\int_0^{\pi/2}\sin^{2m+1}x\,dx = \dfrac{(2m)!!}{(2m+1)!!}$ are sometimes written using double factorial notation in Extension 1 extension problems.

On $[0, \pi/2]$, the reduction $I_n = \dfrac{n-1}{n}\,I_{n-2}$ unwinds like a telescope. For even $n$, chain down to $I_0 = \dfrac{\pi}{2}$; for odd $n$, chain down to $I_1 = 1$.

Pause — copy the Wallis-type closed forms on $[0,\pi/2]$: for even $n$, $I_n=\frac{(n-1)!!}{n!!}\cdot\frac{\pi}{2}$; for odd $n$, $I_n=\frac{(n-1)!!}{n!!}$; verify with $n=2,3$ into your book.

Did you get this? True or false: $\displaystyle\int_0^{\pi/2}\sin^3 x\,dx = \dfrac{2}{3}$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · EVALUATE $I_6$

Evaluate $\displaystyle\int_0^{\pi/2}\sin^6 x\,dx$.

Apply $I_n = \dfrac{n-1}{n}\,I_{n-2}$ repeatedly: $I_6 = \dfrac{5}{6}\,I_4$

On $[0,\pi/2]$ the boundary term vanishes, so the recurrence is clean. $n=6$, so multiply by $\frac{5}{6}$.

PROBLEM 2 · PROVE THE FORMULA

Prove that if $I_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx$, then $n\,I_n = (n-1)\,I_{n-2}$.

Use IBP: $u = \sin^{n-1}x$, $dv = \sin x\,dx$ $\Rightarrow$ $du = (n-1)\sin^{n-2}x\cos x\,dx$, $v = -\cos x$

State the IBP choices clearly — exam markers expect this.

PROBLEM 3 · TANGENT POWER REDUCTION

Given $J_n = \displaystyle\int \tan^n x\,dx$, show that $J_n = \dfrac{\tan^{n-1}x}{n-1} - J_{n-2}$, and use it to find $\displaystyle\int \tan^4 x\,dx$.

Write $\tan^n x = \tan^{n-2}x \cdot \tan^2 x = \tan^{n-2}x(\sec^2 x - 1)$

Use the identity $\tan^2 x = \sec^2 x - 1$ to split the power — this avoids IBP entirely for tangent.

Fill the gap: On $[0,\pi/2]$, $I_4 = \dfrac{3}{4}\cdot\dfrac{1}{2}\cdot$ $= \dfrac{3\pi}{16}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the boundary term in indefinite integrals

The reduction formula $I_n = -\dfrac{\sin^{n-1}x\cos x}{n} + \dfrac{n-1}{n}I_{n-2}$ retains a $-\dfrac{\sin^{n-1}x\cos x}{n}$ term for indefinite integrals. Only on $[0, \pi/2]$ does the boundary term vanish. Dropping it in the indefinite case is a serious error.

Trap 02

Mixing even and odd chains

Even powers chain down to $I_0 = \pi/2$; odd powers chain down to $I_1 = 1$. Accidentally reaching $I_0$ when $n$ is odd (or vice versa) gives a nonsense answer. Always check whether $n$ is even or odd before starting the chain.

Trap 03

Wrong IBP choice for tangent

For $\int\tan^n x\,dx$, do NOT use IBP. Instead split using $\tan^2 x = \sec^2 x - 1$. IBP on tangent powers leads to circular working that never resolves.

Did you get this? True or false: The reduction formula $I_n = \dfrac{n-1}{n}\,I_{n-2}$ holds for ALL integrals of the form $\int\sin^n x\,dx$, not just definite integrals on $[0, \pi/2]$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Use the reduction formula to evaluate $\displaystyle\int_0^{\pi/2}\sin^4 x\,dx$.

Evaluate $\displaystyle\int_0^{\pi/2}\sin^7 x\,dx$ using the chain $I_7 \to I_5 \to I_3 \to I_1$.

Use the reduction formula $J_n = \dfrac{\tan^{n-1}x}{n-1} - J_{n-2}$ to evaluate $\displaystyle\int\tan^3 x\,dx$.

Prove that $\displaystyle\int_0^{\pi/2}\cos^n x\,dx = \int_0^{\pi/2}\sin^n x\,dx$ by using the substitution $x = \dfrac{\pi}{2} - t$.

A student claims $\displaystyle\int_0^{\pi/2}\sin^5 x\,dx = \dfrac{5\pi}{32}$. Find the error and give the correct value.

Odd one out: Three of these results are correct. Which one is NOT?

Revisit your thinking

Earlier you estimated whether $\int_0^{\pi/2}\sin^6 x\,dx$ was larger or smaller than $\dfrac{1}{2}$.

The exact answer is $\dfrac{5\pi}{32} \approx 0.491$ — just under $\dfrac{1}{2}$. The reduction chain was $I_6 = \dfrac{5}{6}\cdot\dfrac{3}{4}\cdot\dfrac{1}{2}\cdot\dfrac{\pi}{2}$. Each successive $I_n$ for higher even powers is smaller, because $\sin^n x \leq \sin^{n-2} x$ on $[0,\pi/2]$. The result converges to 0 as $n \to \infty$.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Evaluate $\displaystyle\int_0^{\pi/2}\sin^4 x\,dx$ using the reduction formula. (2 marks)

auto-saved

UnderstandBand 43 marks

Q2. Prove that if $I_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx$, then $n\,I_n = (n-1)I_{n-2}$. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. Use the reduction formula $J_n = \dfrac{\tan^{n-1}x}{n-1} - J_{n-2}$ to evaluate $\displaystyle\int\tan^5 x\,dx$. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers: 1. $I_4 = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3\pi}{16}$ · 2. $I_7 = \frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}\cdot 1 = \frac{16}{35}$ · 3. $J_3 = \frac{\tan^2 x}{2} - \ln|\cos x| + C$ · 4. Substitute to get $\int_0^{\pi/2}\cos^n t\,dt = I_n$ · 5. Error: used $I_0 = \pi/2$ for odd $n$; correct value is $I_5 = \frac{8}{15}$.

Q1 (2 marks): $I_4 = \frac{3}{4}I_2$ [1]; $I_2 = \frac{1}{2}I_0 = \frac{\pi}{4}$; so $I_4 = \frac{3\pi}{16}$ [1].

Q2 (3 marks): IBP with $u=\sin^{n-1}x$, $dv=\sin x\,dx$ [1]; boundary term = 0, expand using $\cos^2 x = 1 - \sin^2 x$ [1]; collect to get $nI_n = (n-1)I_{n-2}$ [1].

Q3 (3 marks): $J_5 = \frac{\tan^4 x}{4} - J_3$ [1]; $J_3 = \frac{\tan^2 x}{2} - J_1 = \frac{\tan^2 x}{2} - \ln|\sec x|$ [1]; $J_5 = \frac{\tan^4 x}{4} - \frac{\tan^2 x}{2} + \ln|\sec x| + C$ [1].

Boss battle · The Reduction Master

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering reduction formula questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 16 · Mixed Integration Techniques Lesson 18 · Harder Integration Problems →

Module overview · Extension 1 · Checkpoint 4