IB Analysis and Approaches HL · Integral calculus · Section 3 of 13

Integration by parts

3 worked examples · By Álvaro López

Key idea

Integration by parts is the product rule in reverse. It is used to integrate products that substitution cannot solve (such as $x e^{x}$ or $x sin x$ ), and also logarithms and inverse trigonometric functions. In Analysis and Approaches HL it can appear in both papers.

Where the formula comes from

The derivative of a product is $(uv)^{'} = u^{'} v + u v^{'}$ . If you integrate both sides and rearrange, you get:

The idea is to replace a hard integral with an easier one: you choose a factor that simplifies when differentiated ( $u$ ) and another that you know how to integrate ( $\frac{d v}{d x}$ ).

How to choose $u$ and $d v$

A practical rule (usually remembered as ILATE) ranks the types of function by order of preference for $u$ :

Inverse trigonometric functions ( $arctan x$ , $arcsin x$ ).
Logarithms ( $ln x$ ).
Algebraic functions, that is, polynomials ( $x$ , $x^{2}$ ).
Trigonometric functions ( $sin x$ , $cos x$ ).
Exponential functions ( $e^{x}$ , $a^{x}$ ).

Take as $u$ the factor that appears first in the list, and as $d v$ the rest. This is a helpful rule, not a theorem: if the new integral turns out harder after applying it, try the opposite choice.

Cases you should know

A polynomial times an exponential or a trigonometric function. Take $u$ = the polynomial, which simplifies when differentiated. If the polynomial has degree $n$ , you must apply parts $n$ times, repeating the same choice at each step.

A logarithm or an inverse function on its own. As no product is visible, invent one: $ln x = ln x \cdot 1$ , with $u = ln x$ and $\frac{d v}{d x} = 1$ . The same applies to $arctan x$ and $arcsin x$ .

Cyclic integrals. With $e^{a x} sin b x$ or $e^{a x} cos b x$ , applying parts twice brings back the original integral. This is not a mistake: treat it as an unknown and solve for it.

Worked examples

Example 1 Basic

A polynomial times an exponential or a cosine, and $x ln x$

Calculate (with $x > 0$ in part c):

a) $\int x e^{- x} d x$ .

b) $\int x cos (2 x) d x$ .

c) $\int x ln x d x$ .

Solution

a) $u = x$ , $\frac{d v}{d x} = e^{- x}$ ; then $\frac{d u}{d x} = 1$ and $v = - e^{- x}$ : $\int x e^{- x} d x = - x e^{- x} + \int e^{- x} d x = - x e^{- x} - e^{- x} + C = - (x + 1) e^{- x} + C .$

b) $u = x$ , $\frac{d v}{d x} = cos 2 x$ ; $v = \frac{sin 2 x}{2}$ : $\int x cos 2 x d x = \frac{x sin 2 x}{2} - \frac{1}{2} \int sin 2 x d x = \frac{x sin 2 x}{2} + \frac{cos 2 x}{4} + C .$

c) Here the logarithm goes as $u$ (ILATE): $u = ln x$ , $\frac{d v}{d x} = x$ ; $\frac{d u}{d x} = \frac{1}{x}$ and $v = \frac{x ^{2}}{2}$ : $\int x ln x d x = \frac{x ^{2}}{2} ln x - \int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x = \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} + C .$

Example 2 Medium

Repeated parts and an inverse function

a) Calculate $\int x^{2} sin x d x$ .

b) Calculate the exact value of $\int_{0}^{1} arctan x d x$ .

Solution

a) First time: $u = x^{2}$ , $\frac{d v}{d x} = sin x$ ; $\frac{d u}{d x} = 2 x$ and $v = - cos x$ : $\int x^{2} sin x d x = - x^{2} cos x + 2 \int x cos x d x .$ Second time (same choice: the polynomial is $u$ ): $u = x$ , $\frac{d v}{d x} = cos x$ ; $v = sin x$ : $\int x cos x d x = x sin x - \int sin x d x = x sin x + cos x .$ Putting it together: $\int x^{2} sin x d x = - x^{2} cos x + 2 x sin x + 2 cos x + C$ .

b) Take $u = arctan x$ , $\frac{d v}{d x} = 1$ ; $\frac{d u}{d x} = \frac{1}{1 + x ^{2}}$ and $v = x$ : $\int arctan x d x = x arctan x - \int \frac{x}{1 + x ^{2}} d x = x arctan x - \frac{1}{2} ln (1 + x^{2}) + C .$ Then $\int_{0}^{1} arctan x d x = [x arctan x - \frac{1}{2} ln (1 + x^{2})]_{0}^{1} = \frac{π}{4} - \frac{1}{2} ln 2.$

Example 3 Hard

A cyclic integral with limits

a) Find $I = \int e^{2 x} sin x d x$ .

b) Calculate the exact value of $\int_{0}^{π} e^{2 x} sin x d x$ .

Solution

a) First application: $u = sin x$ , $\frac{d v}{d x} = e^{2 x}$ ; $\frac{d u}{d x} = cos x$ , $v = \frac{1}{2} e^{2 x}$ : $I = \frac{1}{2} e^{2 x} sin x - \frac{1}{2} \int e^{2 x} cos x d x .$ Second application, with the same choice (the trigonometric function is $u$ ): $u = cos x$ , $\frac{d v}{d x} = e^{2 x}$ ; $\frac{d u}{d x} = - sin x$ : $\int e^{2 x} cos x d x = \frac{1}{2} e^{2 x} cos x + \frac{1}{2} \int e^{2 x} sin x d x = \frac{1}{2} e^{2 x} cos x + \frac{1}{2} I .$ Substituting: $I = \frac{1}{2} e^{2 x} sin x - \frac{1}{4} e^{2 x} cos x - \frac{1}{4} I .$ Move $\frac{1}{4} I$ to the other side: $\frac{5}{4} I = \frac{1}{4} e^{2 x} (2 sin x - cos x)$ , hence $I = \frac{e ^{2 x} ( 2 sin x - cos x )}{5} + C .$

b) By the previous part: $\int_{0}^{π} e^{2 x} sin x d x = [\frac{e ^{2 x} ( 2 sin x - cos x )}{5}]_{0}^{π} = \frac{e ^{2 π} \cdot 1}{5} - \frac{- 1}{5} = \frac{e ^{2 π} + 1}{5} .$

In the exam: in a cyclic integral, marks are awarded for recognising that $I$ reappears and solving for it. Do not change the choice of $u$ in the second application: if you do, you undo the first and return to the starting point.

Common mistakes

Summary

Formula: $\int u d v = uv - \int v d u$ (the product rule in reverse).
Choose $u$ using ILATE: inverse functions, logarithms, polynomials, trigonometric functions, exponentials.
A polynomial of degree $n$ requires $n$ applications; $ln x$ and $arctan x$ are solved with $d v = d x$ .
In cyclic integrals ( $e^{a x} sin b x$ ), the integral reappears and you solve for it; use the same choice both times.
With limits, evaluate $[uv]$ between them and subtract the remaining definite integral.

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Integration by parts

Where the formula comes from

How to choose u and dv

Cases you should know

Worked examples

Common mistakes

Summary

Want to work on it together?

How to choose $u$ and $d v$