A reduction formula expresses an integral $I_n$ in terms of a lower-index integral $I_{n-1}$ or $I_{n-2}$. The recursive structure lets you bring $\int \sin^{10} x\,dx$ down to a base case you can integrate by inspection. This lesson teaches the two derivation patterns, integration by parts with a clever split, and how to use a derived formula to evaluate a specific case.
+5 XP warm-up Write the integration-by-parts formula.
Write the integration-by-parts formula. Then think: if $I_n = \int \sin^n x\,dx$, what would you choose as $u$ and $dv$ to introduce an $I_{n-2}$ on the right-hand side? Before checking what identity will convert $\sin^{n-2}x \cos^2 x$ back into a $\sin^n$ term? Sketch your reasoning below.
+5 XP to read Every reduction-formula derivation rewards two habits: split the integrand to expose an integrable factor, then...
Every reduction-formula derivation rewards two habits: split the integrand to expose an integrable factor, then use a Pythagorean (or other) identity to fold the result back into $I_n$. The Pythagorean identity turns an off-power expression into a copy of $I_n$, which you then solve for algebraically.
The split-parts-fold flow: (1) split $\sin^n x = \sin^{n-1}x \cdot \sin x$ (or $x^n e^x = x^n \cdot e^x$), (2) integrate by parts with $u, dv$ chosen so $v$ is easy, (3) use $\cos^2 x = 1 - \sin^2 x$ (or whatever identity applies) to fold the leftover back into $I_n$ and solve.
$I_n = -\dfrac{\sin^{n-1}x\,\cos x}{n} + \dfrac{n-1}{n} I_{n-2}$ · $J_n = x^n e^x - n J_{n-1}$
Let $\displaystyle I_n = \int \sin^n x\,dx$ for $n \geq 2$.
Let $\displaystyle I_n = \int \sin^n x\,dx$ for $n \geq 2$. Write $\sin^n x = \sin^{n-1}x \cdot \sin x$ and integrate by parts with
$u = \sin^{n-1}x \Rightarrow du = (n-1)\sin^{n-2}x \cos x\,dx$, $dv = \sin x\,dx \Rightarrow v = -\cos x$.
Then $\displaystyle I_n = -\sin^{n-1}x \cos x + \int (n-1)\sin^{n-2}x \cos^2 x\,dx$.
Use $\cos^2 x = 1 - \sin^2 x$ to split the integral:
$\displaystyle I_n = -\sin^{n-1}x \cos x + (n-1)\int \sin^{n-2}x\,dx - (n-1)\int \sin^n x\,dx$.
The last integral is $I_n$ again, so $I_n + (n-1)I_n = -\sin^{n-1}x \cos x + (n-1)I_{n-2}$, giving
$I_n = \int \sin^n x\,dx$. Split $\sin^n = \sin^{n-1} \cdot \sin x$ · Parts: $u = \sin^{n-1}x$, $dv = \sin x\,dx \Rightarrow v = -\cos x$ · Use $\cos^2 = 1 - \sin^2$ to expose $I_n$ on the RHS · Final: $I_n = -\dfrac{\sin^{n-1}x\,\cos x}{n} + \dfrac{n-1}{n}\,I_{n-2}$
Pause, copy the $\sin^n x$ reduction formula $I_n = -\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}I_{n-2}$, the derivation steps (split, IBP, $\cos^2=1-\sin^2$, expose $I_n$) into your book.
Quick check: In deriving the reduction formula for $I_n = \int \sin^n x\,dx$, which substitution in integration by parts is correct?
We just saw the $\sin^n x$ reduction formula $I_n = -\tfrac{\sin^{n-1}x\cos x}{n} + \tfrac{n-1}{n}I_{n-2}$, derived by splitting off one $\sin x$ as $dv$, applying IBP, then using $\cos^2 = 1-\sin^2$ to expose $I_n$ on the right.
We just saw the $\sin^n x$ reduction formula $I_n = -\tfrac{\sin^{n-1}x\cos x}{n} + \tfrac{n-1}{n}I_{n-2}$, derived by splitting off one $\sin x$ as $dv$, applying IBP, then using $\cos^2 = 1-\sin^2$ to expose $I_n$ on the right. That raises a question: does the same split-off-and-IBP idea work for $\int x^n e^x\,dx$? This card answers it → yes: $u = x^n$, $dv = e^x dx$ gives $J_n = x^n e^x - nJ_{n-1}$, reducing the power by 1 each step.
Let $\displaystyle J_n = \int x^n e^x\,dx$ for $n \geq 1$. Integrate by parts with
$u = x^n \Rightarrow du = n x^{n-1}\,dx$, $dv = e^x\,dx \Rightarrow v = e^x$.
Then $\displaystyle J_n = x^n e^x - n\int x^{n-1} e^x\,dx = x^n e^x - n J_{n-1}$.
Index drops by 1 each iteration. Base case: $J_0 = \int e^x\,dx = e^x + C$.
$J_n = \int x^n e^x\,dx$. Parts: $u = x^n$, $dv = e^x dx \Rightarrow v = e^x$ · $J_n = x^n e^x - n J_{n-1}$. Index drops by 1 · Base case: $J_0 = e^x + C$ · Iterate: $J_2 = x^2 e^x - 2 J_1 = x^2 e^x - 2(x e^x - J_0) = (x^2 - 2x + 2)e^x + C$
Pause, copy $J_n = \int x^n e^x\,dx$, the reduction $J_n = x^n e^x - nJ_{n-1}$, the base $J_0 = e^x+C$, and the iterated example $J_2 = (x^2-2x+2)e^x+C$ into your book.
Did you get this? True or false: the reduction formula $J_n = x^n e^x - n J_{n-1}$ for $\int x^n e^x\,dx$ requires the Pythagorean identity in its derivation.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Show that if $I_n = \displaystyle\int \sin^n x\,dx$, then $I_n = -\dfrac{\sin^{n-1}x\,\cos x}{n} + \dfrac{n-1}{n}\,I_{n-2}$ for $n \geq 2$.
Using the reduction formula from Problem 1, evaluate $I_5 = \displaystyle\int \sin^5 x\,dx$ by reducing through $I_3$ to $I_1$.
Derive the reduction formula for $J_n = \displaystyle\int x^n e^x\,dx$ and use it to find $J_3$.
Fill the gap: The $\sin^n$ reduction formula reads $I_n = -\dfrac{\sin^{n-1}x\,\cos x}{p} + \dfrac{n-1}{n}\,I_{q}$, with base case $I_0 = x + C$ when $n$ is even. Give $p =$ and $q =$ .
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: applying the $\sin^n$ reduction formula starting from $n = 6$ terminates at the base case $I_0 = x + C$.
Activities · practice with the ideas
Activities · practice with the ideas
Use the reduction formula $I_n = -\tfrac{\sin^{n-1}x \cos x}{n} + \tfrac{n-1}{n}I_{n-2}$ to compute $I_4 = \int \sin^4 x\,dx$.
Derive a reduction formula for $C_n = \int \cos^n x\,dx$. State the analogous result.
Apply $J_n = x^n e^x - n J_{n-1}$ to evaluate $J_2 = \int x^2 e^x\,dx$.
Use the definite version: show that $\displaystyle\int_0^{\pi/2} \sin^n x\,dx = \dfrac{n-1}{n}\int_0^{\pi/2} \sin^{n-2}x\,dx$ for $n \geq 2$. (The boundary term vanishes.)
Hence evaluate $\displaystyle\int_0^{\pi/2} \sin^4 x\,dx$.
Odd one out: Three of these are standard reduction formulas. Which is NOT a valid reduction (the recurrence does not lower the index)?
Earlier you wondered whether $\int \sin^7 x\,dx$ could be brought down to $\int \sin^5 x\,dx$, then $\sin^3$, then $\sin$.
The split-parts-fold method does exactly that: writing $\sin^n x = \sin^{n-1}x \cdot \sin x$ and applying integration by parts, then using $\cos^2 x = 1 - \sin^2 x$, exposes a copy of $I_n$ on the right which folds back into the formula. Solving algebraically gives $I_n = -\tfrac{\sin^{n-1}x \cos x}{n} + \tfrac{n-1}{n}I_{n-2}$. The same pattern, split, parts, identity, fold, works for $\cos^n$, $\tan^n$, and many others. For $x^n e^x$ the fold is unnecessary because $e^x$ self-replicates under integration.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Given $J_n = \int x^n e^x\,dx = x^n e^x - n J_{n-1}$, evaluate $J_2$. (2 marks)
Q2. Derive the reduction formula $I_n = -\dfrac{\sin^{n-1}x \cos x}{n} + \dfrac{n-1}{n}I_{n-2}$ for $I_n = \int \sin^n x\,dx$ (with $n \geq 2$). State each algebraic step clearly. (3 marks)
Q3. Hence (using the formula from Q2) evaluate $\displaystyle\int_0^{\pi/2} \sin^5 x\,dx$. (3 marks)
Activity answers:
1. $I_4 = -\dfrac{\sin^3 x \cos x}{4} + \dfrac{3}{4}I_2$; $I_2 = -\dfrac{\sin x \cos x}{2} + \dfrac{1}{2}I_0 = -\dfrac{\sin x \cos x}{2} + \dfrac{x}{2}$. Therefore $I_4 = -\dfrac{\sin^3 x \cos x}{4} - \dfrac{3 \sin x \cos x}{8} + \dfrac{3x}{8} + C$.
2. Split $\cos^n x = \cos^{n-1}x \cdot \cos x$. Parts: $u = \cos^{n-1}x$, $dv = \cos x\,dx$, $v = \sin x$. Use $\sin^2 x = 1 - \cos^2 x$. Result: $C_n = \dfrac{\cos^{n-1}x \sin x}{n} + \dfrac{n-1}{n}C_{n-2}$.
3. $J_0 = e^x + C$; $J_1 = x e^x - J_0 = (x-1)e^x$; $J_2 = x^2 e^x - 2 J_1 = x^2 e^x - 2(x-1)e^x = (x^2 - 2x + 2)e^x + C$.
4. Boundary term $\left[-\dfrac{\sin^{n-1}x \cos x}{n}\right]_0^{\pi/2}$: at $\pi/2$, $\cos(\pi/2) = 0$; at $0$, $\sin^{n-1}(0) = 0$. Both ends zero, so the boundary vanishes, leaving the recurrence.
5. $\displaystyle\int_0^{\pi/2} \sin^4 x\,dx = \dfrac{3}{4} \int_0^{\pi/2} \sin^2 x\,dx = \dfrac{3}{4} \cdot \dfrac{1}{2} \int_0^{\pi/2} 1\,dx = \dfrac{3}{4} \cdot \dfrac{1}{2} \cdot \dfrac{\pi}{2} = \dfrac{3\pi}{16}$.
Q1 (2 marks): $J_1 = xe^x - J_0 = (x-1)e^x$ [1]. $J_2 = x^2 e^x - 2(x-1)e^x = (x^2 - 2x + 2)e^x + C$ [1].
Q2 (3 marks): Setup (split + parts choices) [1]; apply parts and substitute $\cos^2 = 1 - \sin^2$ [1]; collect $I_n$ terms and divide by $n$ [1].
Q3 (3 marks): Boundary term vanishes [1]; apply reduction twice [1]: $\dfrac{4}{5} \cdot \dfrac{2}{3} \cdot \int_0^{\pi/2} \sin x\,dx = \dfrac{8}{15} \cdot 1 = \dfrac{8}{15}$ [1].
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