Integration by Parts: Repeated Application

At A-Level you meet the basic integration-by-parts formula. In Further Mathematics you must apply it repeatedly — twice, three times, or more — and master cyclic integrals, where the original integral reappears and is then solved algebraically rather than by further integration. This lesson develops the repeated technique, the tabular (DI) method for speed, the cyclic trick for $e^{ax}$eax times trig, and the powers-of-logarithm case, all with full mark schemes and exam technique.

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1 Spec mapping (AQA 7367)

Repeated integration by parts is compulsory Pure content for Papers 1 and 2 (each 2 h, 100 marks, 33⅓%, with per-paper weighting AO1 55% / AO2 25% / AO3 20%), within "Further calculus". It is mostly AO1 (carry out a multi-stage technique accurately) with AO2 appearing in "show that" derivations of the cyclic-integral results and in laying out reasoning cleanly, and occasionally AO3 when embedded in a longer modelling problem. It is the engine behind reduction formulae (the next lesson), powers the $(\ln y)^2$(lny)2 volume integral, and reappears in solving second-order differential equations and in Laplace transforms beyond A-Level. Because the calculations are multi-stage, this is a topic where neat, fully-shown working is rewarded as much as the final answer — a tidy layout of $u, dv, du, v$u,dv,du,v at each stage both reduces errors and earns method marks.

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2 Core theory — the formula and choosing $u$u

The integration-by-parts formula, from the product rule, is

$\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx, \qquad\text{shorthand}\quad \int u\,dv = uv - \int v\,du.$∫udxdv​dx=uv−∫vdxdu​dx,shorthand∫udv=uv−∫vdu.

The whole art is choosing $u$u so that $\int v\,du$∫vdu is simpler than $\int u\,dv$∫udv. The LIATE priority for $u$u — Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — picks $u$u as the function earliest in that list (it differentiates toward something simpler), leaving the rest as $dv$dv. For $\int x^n e^{ax}\,dx$∫xneaxdx, the algebraic $x^n$xn is $u$u (it eventually differentiates to $0$0); for $\int x^n \ln x\,dx$∫xnlnxdx, the logarithm $\ln x$lnx is $u$u (because differentiating $\ln x$lnx removes it, whereas integrating it would be awkward).

The reasoning behind LIATE is worth internalising rather than memorising blindly. Logarithms and inverse-trig functions are placed first because they are hard to integrate but easy to differentiate, so we want them as $u$u. Exponentials and trig functions are placed last because they integrate and differentiate equally easily, so they make ideal $dv$dv choices that do not get worse when integrated. Algebraic powers $x^n$xn sit in the middle: differentiating them lowers the power toward $0$0, which is exactly what we want when the other factor is an exponential or trig function. A bad choice of $u$u makes $\int v\,du$∫vdu harder than the original — for instance taking $u = e^x$u=ex in $\int x e^x\,dx$∫xexdx leads to $\int \tfrac{x^2}{2}e^x\,dx$∫2x2​exdx, a step backwards. When in doubt, try LIATE; if the resulting integral looks worse, swap your choice.

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3 Worked examples with M1/A1 mark scheme

Worked Example 1 — parts applied twice

Evaluate $\displaystyle \int x^2 e^{x}\,dx.$∫x2exdx.

First application ($u=x^2,\ dv=e^x dx \Rightarrow du=2x\,dx,\ v=e^x$u=x2, dv=exdx⇒du=2xdx, v=ex):

$\int x^2 e^{x}\,dx = x^2 e^{x} - \int 2x e^{x}\,dx \qquad\text{(M1 one correct application)}$∫x2exdx=x2ex−∫2xexdx(M1 one correct application)

Second application on $\int 2x e^x\,dx$∫2xexdx ($u=2x,\ dv=e^x dx$u=2x, dv=exdx):

$\int 2x e^{x}\,dx = 2x e^{x} - \int 2 e^{x}\,dx = 2x e^{x} - 2e^{x} + C.$∫2xexdx=2xex−∫2exdx=2xex−2ex+C.

Combining: (M1 second application; A1 final form)

$\int x^2 e^{x}\,dx = e^{x}\big(x^2 - 2x + 2\big) + C.$∫x2exdx=ex(x2−2x+2)+C.

(Check by differentiating: $\tfrac{d}{dx}\big[e^x(x^2-2x+2)\big] = e^x(x^2-2x+2) + e^x(2x-2) = e^x\big[(x^2-2x+2)+(2x-2)\big] = e^x x^2$dxd​[ex(x2−2x+2)]=ex(x2−2x+2)+ex(2x−2)=ex[(x2−2x+2)+(2x−2)]=exx2. ✓) The pattern here is worth noticing: $\int x^2 e^x\,dx = e^x(x^2 - 2x + 2)$∫x2exdx=ex(x2−2x+2), and more generally $\int x^n e^x\,dx$∫xnexdx produces $e^x$ex times an alternating polynomial $x^n - n x^{n-1} + n(n-1)x^{n-2} - \dots$xn−nxn−1+n(n−1)xn−2−…. The tabular method (next section) makes this pattern transparent and is the fastest way to obtain it for larger $n$n.

Worked Example 2 — parts applied three times

Evaluate $\displaystyle \int x^3 \sin x\,dx.$∫x3sinxdx. Apply parts three times, reducing the power of $x$x by one at each stage until it disappears:

$\begin{aligned} \int x^3\sin x\,dx &= -x^3\cos x + \int 3x^2\cos x\,dx \\ &= -x^3\cos x + 3x^2\sin x - \int 6x\sin x\,dx \\ &= -x^3\cos x + 3x^2\sin x + 6x\cos x - \int 6\cos x\,dx \\ &= -x^3\cos x + 3x^2\sin x + 6x\cos x - 6\sin x + C. \end{aligned}$∫x3sinxdx​=−x3cosx+∫3x2cosxdx=−x3cosx+3x2sinx−∫6xsinxdx=−x3cosx+3x2sinx+6xcosx−∫6cosxdx=−x3cosx+3x2sinx+6xcosx−6sinx+C.​

(M1 for each of the three applications, A1 final answer.) Check by differentiating recovers $x^3\sin x$x3sinx. ✓

Worked Example 2b — a logarithm-only integrand

Evaluate $\displaystyle \int \ln x\,dx.$∫lnxdx. Although there is no obvious product, integration by parts handles it by treating the integrand as $\ln x \times 1$lnx×1, with $u = \ln x$u=lnx and $dv = 1\,dx$dv=1dx (so $du = \tfrac1x\,dx$du=x1​dx, $v = x$v=x):

$\int \ln x\,dx = x\ln x - \int x\cdot\frac1x\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.$∫lnxdx=xlnx−∫x⋅x1​dx=xlnx−∫1dx=xlnx−x+C.

This "$dv = 1\,dx$dv=1dx" trick is the standard way to integrate $\ln x$lnx, $\arctan x$arctanx and other functions that are hard to integrate but easy to differentiate — and it is the base case for the $(\ln x)^n$(lnx)n family that appears in volumes of revolution about the $y$y-axis.

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4 The tabular (DI) method

For $x^n$xn multiplied by $e^{ax}$eax, $\sin(ax)$sin(ax) or $\cos(ax)$cos(ax), build two columns: Differentiate the polynomial down to $0$0; Integrate the other factor repeatedly. Attach alternating signs $+,-,+,-,\dots$+,−,+,−,… and multiply along the diagonals (each row's $D$D-entry with the next row's $I$I-entry). The method is especially valuable when $n$n is $3$3 or more, where doing parts line-by-line three or four times invites sign slips; the table keeps every term and its sign organised in one place.

Tabular example: $\displaystyle \int x^2 e^{3x}\,dx$∫x2e3xdx

Sign	$D$D: derivatives of $x^2$x2	$I$I: integrals of $e^{3x}$e3x
$+$+	$x^2$x2	$e^{3x}$e3x
$-$−	$2x$2x	$\tfrac13 e^{3x}$31​e3x
$+$+	$2$2	$\tfrac19 e^{3x}$91​e3x
$-$−	$0$0	$\tfrac{1}{27} e^{3x}$271​e3x

$\int x^2 e^{3x}\,dx = (+)\,x^2\cdot\tfrac13 e^{3x} + (-)\,2x\cdot\tfrac19 e^{3x} + (+)\,2\cdot\tfrac{1}{27} e^{3x} + C = \left(\frac{x^2}{3} - \frac{2x}{9} + \frac{2}{27}\right)e^{3x} + C,$∫x2e3xdx=(+)x2⋅31​e3x+(−)2x⋅91​e3x+(+)2⋅271​e3x+C=(3x2​−92x​+272​)e3x+C,

which tidies to $\dfrac{e^{3x}}{27}\big(9x^2 - 6x + 2\big) + C.$27e3x​(9x2−6x+2)+C.

The tabular method is simply repeated integration by parts organised into a table, so it gives identical results — but with far less chance of a sign or bookkeeping error when the polynomial power is high. It works whenever one factor differentiates to $0$0 after finitely many steps (any polynomial $x^n$xn) and the other can be integrated repeatedly ($e^{ax}$eax, $\sin ax$sinax, $\cos ax$cosax). The two rules to remember: attach alternating signs $+,-,+,-,\dots$+,−,+,−,… starting with $+$+ on the top row, and multiply each $D$D-entry by the $I$I-entry one row down (the diagonal). Stop when the derivative column reaches $0$0; the last non-zero product is the final term. For a definite integral, write down the antiderivative from the table first, then evaluate at the limits — do not try to put limits inside the table itself.

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5 Cyclic integrals (the key Further-Maths trick)

When the integrand is $e^{ax}$eax times $\sin bx$sinbx or $\cos bx$cosbx, parts never terminates — instead the original integral reappears, and you solve for it algebraically.

Worked Example 3 — the canonical cyclic integral

Evaluate $I = \displaystyle \int e^{x}\cos x\,dx.$I=∫excosxdx.

First ($u=\cos x,\ dv=e^x dx$u=cosx, dv=exdx): $\;I = e^{x}\cos x + \displaystyle\int e^{x}\sin x\,dx.$I=excosx+∫exsinxdx.

Second on $\int e^x\sin x\,dx$∫exsinxdx ($u=\sin x,\ dv=e^x dx$u=sinx, dv=exdx):

$I = e^{x}\cos x + e^{x}\sin x - \int e^{x}\cos x\,dx = e^{x}\cos x + e^{x}\sin x - I.$I=excosx+exsinx−∫excosxdx=excosx+exsinx−I.

The last integral is $I$I itself, so $2I = e^{x}(\cos x + \sin x)$2I=ex(cosx+sinx), giving

$I = \frac{e^{x}}{2}\big(\cos x + \sin x\big) + C.$I=2ex​(cosx+sinx)+C.

Key technique. When the original integral reappears, name it $I$I, collect it on one side, and solve. The constant $+C$+C is introduced after dividing.

Two subtleties make or break a cyclic integral. First, consistency of choice. Across the two applications of parts you must keep choosing $dv = e^{x}\,dx$dv=exdx (or consistently the trig factor) — if you swap which factor is $u$u halfway through, the integral "unwinds" back to where it started and you get the useless identity $I = I$I=I. The rule is: pick $e^{x}\,dx$exdx (or $e^{ax}\,dx$eaxdx) as $dv$dv both times, or pick the trig factor as $dv$dv both times. Second, the sign that produces the reappearance. It is the combination of two minus signs from two applications of parts that flips the reappearing integral to $-I$−I (rather than $+I$+I); if you find the integral coming back as $+I$+I, a sign has gone astray and the method appears to fail. Done correctly, you always reach an equation like $I = (\text{algebraic terms}) - I$I=(algebraic terms)−I or $\lambda I = (\text{algebraic terms})$λI=(algebraic terms) for some constant $\lambda$λ, which you solve in one line.

Worked Example 4 — different coefficients

Evaluate $J = \displaystyle \int e^{2x}\sin 3x\,dx.$J=∫e2xsin3xdx.

First ($u=\sin 3x,\ dv=e^{2x}dx,\ v=\tfrac12 e^{2x}$u=sin3x, dv=e2xdx, v=21​e2x): $\;J = \tfrac12 e^{2x}\sin 3x - \tfrac32\displaystyle\int e^{2x}\cos 3x\,dx.$J=21​e2xsin3x−23​∫e2xcos3xdx.

Second ($u=\cos 3x,\ v=\tfrac12 e^{2x}$u=cos3x, v=21​e2x — note we keep $dv = e^{2x}\,dx$dv=e2xdx, the same choice as the first application, which is what makes the original integral reappear):

$J = \tfrac12 e^{2x}\sin 3x - \tfrac32\!\left[ \tfrac12 e^{2x}\cos 3x + \tfrac32\!\int e^{2x}\sin 3x\,dx \right] = \tfrac12 e^{2x}\sin 3x - \tfrac34 e^{2x}\cos 3x - \tfrac94 J.$J=21​e2xsin3x−23​[21​e2xcos3x+23​∫e2xsin3xdx]=21​e2xsin3x−43​e2xcos3x−49​J.

$J + \tfrac94 J = \tfrac12 e^{2x}\sin 3x - \tfrac34 e^{2x}\cos 3x \;\Rightarrow\; \tfrac{13}{4}J = \tfrac{e^{2x}}{4}\big(2\sin 3x - 3\cos 3x\big),$J+49​J=21​e2xsin3x−43​e2xcos3x⇒413​J=4e2x​(2sin3x−3cos3x),

$J = \frac{e^{2x}}{13}\big(2\sin 3x - 3\cos 3x\big) + C.$J=13e2x​(2sin3x−3cos3x)+C.

This matches the general results, which you can quote (for "evaluate") or derive by differentiation (for "show that"):

$\int e^{ax}\sin bx\,dx = \frac{e^{ax}\big(a\sin bx - b\cos bx\big)}{a^2 + b^2} + C, \quad \int e^{ax}\cos bx\,dx = \frac{e^{ax}\big(a\cos bx + b\sin bx\big)}{a^2 + b^2} + C.$∫eaxsinbxdx=a2+b2eax(asinbx−bcosbx)​+C,∫eaxcosbxdx=a2+b2eax(acosbx+bsinbx)​+C.

To verify the first by differentiation, apply the product rule to $\dfrac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2}$a2+b2eax(asinbx−bcosbx)​:

$\frac{1}{a^2+b^2}\Big[ a e^{ax}(a\sin bx - b\cos bx) + e^{ax}(ab\cos bx + b^2\sin bx) \Big] = \frac{e^{ax}\big[(a^2+b^2)\sin bx\big]}{a^2+b^2} = e^{ax}\sin bx,$a2+b21​[aeax(asinbx−bcosbx)+eax(abcosbx+b2sinbx)]=a2+b2eax[(a2+b2)sinbx]​=eaxsinbx,

since the $\cos bx$cosbx terms $-ab$−ab and $+ab$+ab cancel and the $\sin bx$sinbx terms combine to $(a^2+b^2)$(a2+b2). The denominator cancels, recovering the integrand exactly. This verification is an acceptable derivation when a question says "show that the integral equals …", and it is far quicker than two applications of parts — a useful time-saver under exam pressure.

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5a Synoptic links

Repeated integration by parts is one of the most connected techniques in Further calculus. It is the engine of reduction formulae (the next lesson): every reduction formula $I_n = \dots I_{n-2}$In​=…In−2​ is derived by a single, carefully-arranged application of parts. It supplies the $(\ln y)^2$(lny)2 integral needed for volumes of revolution about the $y$y-axis, and the $x\sin x$xsinx integral needed for the shell method. The cyclic case ($e^{ax}$eax times trig) links beautifully to complex numbers via Euler's formula (see Going further), and to second-order differential equations, where particular integrals of the form $e^{ax}\sin bx$eaxsinbx arise constantly. Even improper integrals lean on it — e.g. $\int_0^{\infty} x^n e^{-x}\,dx = n!$∫0∞​xne−xdx=n! is repeated parts plus a limit. Recognising parts as a single tool with many applications, rather than a separate trick per problem type, is what makes the rest of the course cohere.

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6 Specimen-style exam question

(Specimen-style — not a real past paper.)