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Module 15 · L04 of 16 ~45 min ⚡ +90 XP available

Integration by Parts II — Repeated Application

One round of IBP often isn't enough. For $\int x^2 \sin x\,dx$ you need two rounds — the power of $x$ drops each time. For $\int e^x \sin x\,dx$ you loop twice and recover the original integral, then rearrange. This lesson covers both repeated and cyclic IBP, plus the tabular shortcut that turns $\int x^n \times (\text{exp or trig})\,dx$ into a one-line calculation.

Today's hook — Try integration by parts on $\int e^x \sin x\,dx$ once. You'll end up with $\int e^x \cos x\,dx$. Apply IBP again on that. You'll end up with $\int e^x \sin x\,dx$ — the original! Don't panic. Call it $I$, rearrange, and solve algebraically for $I$. That's the cyclic trick.

0/5QUESTS

Recall — your gut answer first

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Recall lesson 03: for $\int x \sin x\,dx$ we chose $u = x$. Now for $\displaystyle \int x^2 \sin x\,dx$ — same LIATE choice, $u = x^2$. After one IBP application, what does the new integral look like? Will another IBP be needed?

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The two patterns of repeated parts

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Repeated IBP comes in two flavours. Reductive: the algebraic factor drops a degree each round until it vanishes ($\int x^n \sin x\,dx$ takes $n$ rounds). Cyclic: the integral reappears after two rounds ($\int e^x \sin x\,dx$), so you set it equal to itself and solve. Recognise which kind you have and you've solved the problem.

The recognise-repeat-rearrange reading: (1) identify whether the integral is reductive (polynomial $\times$ exp/trig) or cyclic (exp $\times$ trig), (2) apply IBP — once for each degree drop, or twice for a cyclic loop, (3) for cyclic integrals, name the integral $I$ and solve algebraically.

Tabular: for $\int x^n \cdot f(x)\,dx$ with $f \in \{e^x, \sin x, \cos x\}$, list derivatives of $x^n$ in one column and antiderivatives of $f$ in the other, multiply diagonally with alternating signs.

$I \;=\; \cdots \;\pm\; I \;\Rightarrow\; \text{solve for } I$

Stick with the same $u$ choice

In the second round, keep the polynomial as $u$ (or the same exp/trig in cyclic problems). Switching mid-stream undoes your first IBP and sends you in circles.

Cyclic: name the integral $I$

When the original integral reappears after two rounds, write $I = (\text{stuff}) - I$ or $I = (\text{stuff}) + I$ and solve. If you get $0 = (\text{stuff})$ the signs are wrong — check.

Tabular: alternating signs $+, -, +, -$

Tabular IBP works whenever differentiating one factor eventually hits zero. Multiply diagonals from each row of the $D$ (derivative) column to the next row of the $I$ (integral) column, with signs $+, -, +, \ldots$.

What you'll master

Know

Key facts

Reductive IBP: $\int x^n \cdot f(x)\,dx$ takes $n$ rounds when $f$ is exp/trig
Cyclic IBP: $\int e^{ax}\sin bx\,dx$, $\int e^{ax}\cos bx\,dx$ loop after two rounds
For cyclic problems: $I = \cdots - I$ then $2I = \cdots$
Tabular IBP encodes repeated reductive IBP as a single signed sum

Understand

Concepts

Why the polynomial degree determines the number of rounds
Why $e^x \sin x$ cycles after two IBP rounds
Why tabular IBP is just bookkeeping for the same calculation

Can do

Skills

Evaluate $\int x^2 \sin x\,dx$, $\int x^2 e^x\,dx$ with two rounds of IBP
Solve $\int e^x \sin x\,dx$ by the loop-and-rearrange method
Apply tabular integration to any $\int x^n \cdot f(x)\,dx$ with $f$ exp or trig

Key terms

Reductive IBPEach application reduces the polynomial degree by 1. For $\int x^n f(x)\,dx$ ($f$ = exp/trig), takes $n$ rounds.

Cyclic (loop) IBPAfter two applications, the original integral reappears. Solve algebraically: $I = \cdots \pm I \Rightarrow 2I = \cdots$.

Tabular integrationShortcut for reductive IBP: list derivatives of polynomial in $D$ column, antiderivatives of $f$ in $I$ column, multiply diagonally with alternating signs.

$D$ column (derivative)Repeated derivatives of the polynomial. Stop when you hit zero.

$I$ column (integral)Repeated antiderivatives of the exp/trig factor. Same number of rows as $D$.

Sign pattern $+, -, +, -$Alternating signs attached to each diagonal product in tabular IBP. Starts with $+$ on the top row.

MEX-C1NESA outcome (Further Integration): applies further integration techniques including repeated integration by parts.

Reductive parts — drop the degree

core concept

For $\int x^n \cdot (\text{exp or trig})\,dx$, each IBP round reduces the power of $x$ by 1. So $\int x^2 \sin x\,dx$ needs two rounds; $\int x^3 e^x\,dx$ needs three. The rule: keep $u$ as the polynomial in every round.

$$\int x^2 \sin x\,dx \;\xrightarrow{\text{IBP}}\; -x^2\cos x + 2\!\int x\cos x\,dx \;\xrightarrow{\text{IBP}}\; -x^2\cos x + 2\big[x\sin x + \cos x\big] + C$$

Why it terminates. Each derivative drops the degree: $x^2 \to 2x \to 2 \to 0$. After three derivatives the polynomial vanishes, ending the recursion.

Bookkeeping tip. Track the running answer in a single expression rather than restarting after each round. The coefficient $2$ in $2\int x \cos x\,dx$ stays with that integral all the way to the end.

Reductive IBP: keep $u$ as the polynomial in every round · $\int x^n f(x)\,dx$ needs $n$ rounds (where $f$ = exp or trig) · $\int x^2 \sin x\,dx = -x^2 \cos x + 2x\sin x + 2\cos x + C$ · Carry coefficients through; don't restart after each round

Pause — copy the reductive-IBP rule (keep polynomial as $u$, $n$ rounds needed), the result $\int x^2\sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x + C$, and the carry-coefficients-through instruction into your book.

Quick check: How many applications of IBP are needed to evaluate $\int x^3 e^x \, dx$?

Cyclic parts — name the integral and solve

core concept

We just saw reductive IBP: keep $u$ as the polynomial each round, letting the degree drop by 1 each time; $\int x^n f(x)\,dx$ needs $n$ rounds. That raises a question: what if IBP reproduces the original integral rather than simplifying it? This card answers it → name the integral $I$, perform two rounds of IBP, recover $I$ on the right, then solve algebraically to get $2I = $ (boundary terms).

For $\int e^{ax}\sin bx\,dx$ or $\int e^{ax}\cos bx\,dx$, the polynomial degree never drops (there isn't one). After two rounds of IBP, the original integral reappears with a coefficient. Then it's algebra, not calculus.

Procedure. Let $I = \int e^x \sin x\,dx$.

Round 1 ($u = \sin x$, $dv = e^x\,dx$): $I = e^x \sin x - \int e^x \cos x\,dx$.
Round 2 on $\int e^x \cos x\,dx$ ($u = \cos x$, $dv = e^x\,dx$): $\int e^x \cos x\,dx = e^x \cos x + \int e^x \sin x\,dx = e^x \cos x + I$.
Substitute back: $I = e^x \sin x - (e^x \cos x + I) = e^x \sin x - e^x \cos x - I$.
Solve: $2I = e^x(\sin x - \cos x) \;\Rightarrow\; I = \tfrac{1}{2}e^x(\sin x - \cos x) + C$.

$$\int e^x \sin x\,dx \;=\; \tfrac{1}{2}\,e^x(\sin x - \cos x) + C$$

Critical step. You must keep the same kind of $u$ in both rounds — both trig, or both exp, or both polynomial. Swapping mid-stream undoes round 1 and you get $I = I$ (no information). The cyclic trick only works when round 2 keeps the same convention as round 1.

Cyclic IBP: name the integral $I$, do two rounds, recover $I$ on the right · Solve algebraically: $I = (\text{stuff}) - I \Rightarrow 2I = (\text{stuff})$ · $\int e^x \sin x\,dx = \tfrac{1}{2}e^x(\sin x - \cos x) + C$ · $\int e^x \cos x\,dx = \tfrac{1}{2}e^x(\sin x + \cos x) + C$ · Keep the same kind of $u$ in both rounds, or the loop fails

Pause — copy the cyclic-IBP procedure (name $I$, two rounds, recover $I$, solve $2I = \ldots$), the results $\int e^x\sin x\,dx = \tfrac{1}{2}e^x(\sin x-\cos x)+C$ and $\int e^x\cos x\,dx = \tfrac{1}{2}e^x(\sin x+\cos x)+C$, into your book.

Did you get this? True or false: when applying repeated IBP to $\int e^x \cos x\,dx$, you must change the type of $u$ between the two rounds to avoid a loop.

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

PROBLEM 1 · $\int x^2 \sin x \, dx$ (REDUCTIVE)

Evaluate $\displaystyle \int x^2 \sin x \, dx$ by applying IBP twice.

Round 1: $u = x^2$, $dv = \sin x\,dx \Rightarrow du = 2x\,dx$, $v = -\cos x$. So $\int x^2 \sin x\,dx = -x^2 \cos x + 2\int x \cos x\,dx$.

LIATE: Algebraic beats Trig. Differentiating $x^2$ drops the degree to 1, setting up the next round.

PROBLEM 2 · $\int e^x \sin x \, dx$ (CYCLIC)

Evaluate $\displaystyle \int e^x \sin x \, dx$ using the loop-and-rearrange method.

Let $I = \int e^x \sin x\,dx$. Round 1: $u = \sin x$, $dv = e^x\,dx \Rightarrow du = \cos x\,dx$, $v = e^x$. So $I = e^x \sin x - \int e^x \cos x\,dx$.

LIATE doesn't help here — neither factor simplifies under differentiation. Choose either; we'll pick $u =$ trig and stay consistent.

PROBLEM 3 · TABULAR INTEGRATION FOR $\int x^3 e^{2x} \, dx$

Evaluate $\displaystyle \int x^3 e^{2x} \, dx$ using the tabular shortcut.

Set up two columns: $D$ (derivatives of $x^3$) and $I$ (antiderivatives of $e^{2x}$). $D$: $x^3,\; 3x^2,\; 6x,\; 6,\; 0$. $I$: $e^{2x},\; \tfrac{1}{2}e^{2x},\; \tfrac{1}{4}e^{2x},\; \tfrac{1}{8}e^{2x},\; \tfrac{1}{16}e^{2x}$.

Keep differentiating until the $D$ column hits zero. Antidifferentiate the same number of times in the $I$ column (each integration of $e^{2x}$ multiplies by $\tfrac{1}{2}$).

Fill the gap: For the cyclic integral $\int e^x \sin x\,dx$ we set $I = \int e^x \sin x\,dx$. After two IBP rounds we get $I = e^x \sin x - e^x \cos x - I$. Solving: $2I = $ , so $I = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Switching $u$ between rounds in cyclic IBP

If you pick $u = \sin x$ in round 1 of $\int e^x \sin x\,dx$, you must pick $u = \cos x$ (still trig) in round 2. Switching to $u = e^x$ in round 2 undoes round 1 and produces $I = I$, telling you nothing. Stay consistent.

Trap 02

Tabular signs starting wrong

The first diagonal product gets a $+$ sign, not $-$. Sign pattern: $+, -, +, -, \ldots$. Off-by-one sign errors flip every term and ruin the answer. Mark the signs in the margin before multiplying.

Trap 03

Tabular with non-polynomial $u$

Tabular IBP only works when the $D$ column eventually hits zero — i.e., when $u$ is a polynomial. Trying to tabularise $\int e^x \sin x\,dx$ fails because neither column terminates. Use the cyclic loop method instead.

Did you get this? True or false: tabular integration is appropriate for evaluating $\int e^x \cos x \, dx$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Evaluate $\displaystyle \int x^2 e^x \, dx$ by applying IBP twice.

Evaluate $\displaystyle \int x^2 \cos x \, dx$ using tabular integration.

Evaluate $\displaystyle \int e^{2x} \cos 3x \, dx$ using the cyclic method. Let $I$ denote the integral.

Evaluate $\displaystyle \int_0^{\pi} x^2 \sin x \, dx$. Apply the result from Worked Example 1.

Derive $\displaystyle \int e^x \cos x \, dx$ using the same loop method as Worked Example 2.

Odd one out: Three of these integrals are reductive (tabular shortcut works). Which one is cyclic (needs the loop-and-rearrange method)?

Revisit your thinking

Earlier you previewed $\int x^2 \sin x\,dx$ and saw that one round of IBP produces another product integral.

The pattern that emerged: every round of IBP on a polynomial $\times$ exp/trig integrand drops the polynomial degree by one. When the polynomial vanishes, the recursion ends. The tabular shortcut compresses this whole calculation into one signed sum. Cyclic integrals like $\int e^x \sin x\,dx$ are different — the integral reappears after two rounds, so you solve algebraically. Recognise the pattern up front (polynomial $\to$ reductive/tabular; exp $\times$ trig $\to$ cyclic) and the technique writes itself.

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Multiple choice

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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 33 marks

Q1. Evaluate $\displaystyle \int x^2 e^x \, dx$ using two applications of IBP. (3 marks)

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ApplyBand 43 marks

Q2. Use tabular integration to evaluate $\displaystyle \int x^3 \sin x \, dx$. Show the $D$ and $I$ columns. (3 marks)

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AnalyseBand 54 marks

Q3. Let $I = \displaystyle \int e^{x} \cos x \, dx$. Apply IBP twice to derive an equation involving $I$, and hence evaluate $I$. (4 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Round 1: $u = x^2$, $dv = e^x\,dx \Rightarrow x^2 e^x - 2\int x e^x\,dx$. Round 2: $\int x e^x\,dx = (x-1)e^x$. Total: $x^2 e^x - 2(x-1)e^x + C = (x^2 - 2x + 2)e^x + C$.

2. Tabular. $D$: $x^2, 2x, 2, 0$. $I$: $\cos x, \sin x, -\cos x, -\sin x$. Signs $+, -, +$. Answer: $x^2 \sin x + 2x\cos x - 2\sin x + C$.

3. Let $I = \int e^{2x}\cos 3x\,dx$. Round 1 ($u = \cos 3x$, $dv = e^{2x}\,dx$): $I = \tfrac{1}{2}e^{2x}\cos 3x + \tfrac{3}{2}\int e^{2x}\sin 3x\,dx$. Round 2 ($u = \sin 3x$): $\int e^{2x}\sin 3x\,dx = \tfrac{1}{2}e^{2x}\sin 3x - \tfrac{3}{2}I$. Substitute: $I = \tfrac{1}{2}e^{2x}\cos 3x + \tfrac{3}{4}e^{2x}\sin 3x - \tfrac{9}{4}I$. So $\tfrac{13}{4}I = \tfrac{1}{4}e^{2x}(2\cos 3x + 3\sin 3x)$, giving $I = \tfrac{1}{13}e^{2x}(2\cos 3x + 3\sin 3x) + C$.

4. From Worked Example 1, antiderivative is $-x^2\cos x + 2x\sin x + 2\cos x$. At $x = \pi$: $-\pi^2(-1) + 2\pi(0) + 2(-1) = \pi^2 - 2$. At $x = 0$: $0 + 0 + 2 = 2$. Definite integral $= (\pi^2 - 2) - 2 = \pi^2 - 4$.

5. $J = \int e^x \cos x\,dx$. Round 1 ($u = \cos x$): $J = e^x \cos x + \int e^x \sin x\,dx$. Round 2 ($u = \sin x$): $\int e^x \sin x\,dx = e^x \sin x - J$. So $J = e^x \cos x + e^x \sin x - J \Rightarrow 2J = e^x(\sin x + \cos x) \Rightarrow J = \tfrac{1}{2}e^x(\sin x + \cos x) + C$.

Q1 (3 marks): Round 1 setup [1]; round 2 $\int x e^x\,dx = (x-1)e^x$ [1]; combine to $(x^2 - 2x + 2)e^x + C$ [1].

Q2 (3 marks): $D$/$I$ columns shown [1]; correct signs $+, -, +, -$ [1]; final answer $-x^3 \cos x + 3x^2 \sin x + 6x\cos x - 6\sin x + C$ [1].

Q3 (4 marks): Round 1 setup [1]; round 2 setup with same convention [1]; substitution yielding $I = e^x \cos x + e^x \sin x - I$ [1]; solve $2I = e^x(\sin x + \cos x) \Rightarrow I = \tfrac{1}{2}e^x(\sin x + \cos x) + C$ [1].

Boss battle · The Tabular Titan

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Five timed questions on repeated and cyclic IBP, plus tabular shortcuts. 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 quick repeated-IBP questions. Lighter alternative to the boss.

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Module overview · Extension 2