Integration by Parts: Repeated Application

At A-Level you learn the basic integration by parts formula. In Further Mathematics, you need to apply it repeatedly — sometimes two, three, or even more times — and handle cyclic integrals where the original integral reappears during the process. This lesson covers these advanced techniques, including the tabular method for speed.

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1 Review of Integration by Parts

The formula is:

integral of u (dv/dx) dx = uv - integral of v (du/dx) dx

Or in shorthand: integral of u dv = uv - integral of v du.

The key skill is choosing u and dv wisely. The LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) suggests a priority order for u.

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2 Repeated Integration by Parts

Sometimes one application of integration by parts does not finish the job — the resulting integral still requires parts. You simply apply the formula again.

Worked Example 1

Evaluate the integral of x^2 e^x dx.

Solution:

First application: u = x^2, dv = e^x dx. du = 2x dx, v = e^x.

integral of x^2 e^x dx = x^2 e^x - integral of 2x e^x dx

Second application (on integral of 2x e^x dx): u = 2x, dv = e^x dx. du = 2 dx, v = e^x.

integral of 2x e^x dx = 2x e^x - integral of 2 e^x dx = 2x e^x - 2e^x + C

Combining:

integral of x^2 e^x dx = x^2 e^x - 2x e^x + 2e^x + C = e^x(x^2 - 2x + 2) + C

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Worked Example 2

Evaluate the integral of x^3 sin x dx.

Solution:

This requires three applications of integration by parts.

First: u = x^3, dv = sin x dx. Then du = 3x^2 dx, v = -cos x.

= -x^3 cos x + integral of 3x^2 cos x dx

Second: u = 3x^2, dv = cos x dx. Then du = 6x dx, v = sin x.

= -x^3 cos x + 3x^2 sin x - integral of 6x sin x dx

Third: u = 6x, dv = sin x dx. Then du = 6 dx, v = -cos x.

= -x^3 cos x + 3x^2 sin x + 6x cos x - integral of 6 cos x dx

= -x^3 cos x + 3x^2 sin x + 6x cos x - 6 sin x + C

integral of x^3 sin x dx = -x^3 cos x + 3x^2 sin x + 6x cos x - 6 sin x + C

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3 The Tabular Method

When integrating x^n multiplied by e^(ax), sin(ax), or cos(ax), the tabular method provides a systematic shortcut.

How It Works

Create two columns:

• Left column (u and its derivatives): Start with the polynomial and differentiate repeatedly until you reach zero.
• Right column (dv and its integrals): Start with the other function and integrate repeatedly.

Then alternate signs (+, -, +, -, ...) and multiply diagonally (left entry times the next right entry).

Tabular Example: integral of x^2 e^(3x) dx

Sign	Derivatives of x^2	Integrals of e^(3x)
+	x^2	e^(3x)
-	2x	(1/3) e^(3x)
+	2	(1/9) e^(3x)
-	0	(1/27) e^(3x)

Multiply diagonally with alternating signs:

= (+) x^2 * (1/3) e^(3x) + (-) 2x * (1/9) e^(3x) + (+) 2 * (1/27) e^(3x) + C

= (x^2/3 - 2x/9 + 2/27) e^(3x) + C

= (e^(3x)/27)(9x^2 - 6x + 2) + C

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4 Cyclic Integrals

A cyclic integral occurs when repeated integration by parts brings back the original integral. This happens typically with products of exponential and trigonometric functions.

Worked Example 3

Evaluate I = integral of e^x cos x dx.

Solution:

First application: u = cos x, dv = e^x dx. du = -sin x dx, v = e^x.

I = e^x cos x + integral of e^x sin x dx

Second application (on integral of e^x sin x dx): u = sin x, dv = e^x dx. du = cos x dx, v = e^x.

I = e^x cos x + e^x sin x - integral of e^x cos x dx

But the last integral is just I again! So:

I = e^x cos x + e^x sin x - I

2I = e^x cos x + e^x sin x

I = (e^x/2)(cos x + sin x) + C