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.
+5 XP warm-up Recall lesson 03: for $\int x \sin x\,dx$ we chose $u = x$.
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?
+5 XP to read Repeated IBP comes in two flavours.
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.
For $\int x^n \cdot (\text{exp or trig})\,dx$, each IBP round reduces the power of $x$ by 1.
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.
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.
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 $\int x^3 e^x \, dx$?
We just saw reductive IBP: keep as the polynomial each round, letting the degree drop by 1 each time; needs rounds.
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 $I = \int e^x \sin x\,dx$.
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
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle \int x^2 \sin x \, dx$ by applying IBP twice.
Evaluate $\displaystyle \int e^x \sin x \, dx$ using the loop-and-rearrange method.
Evaluate $\displaystyle \int x^3 e^{2x} \, dx$ using the tabular shortcut.
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
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: tabular integration is appropriate for evaluating $\int e^x \cos x \, dx$.
Activities · practice with the ideas
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)?
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.
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. Evaluate $\displaystyle \int x^2 e^x \, dx$ using two applications of IBP. (3 marks)
Q2. Use tabular integration to evaluate $\displaystyle \int x^3 \sin x \, dx$. Show the $D$ and $I$ columns. (3 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)
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].
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