Integration by Parts Revision Notes for Leaving Cert Applied Maths

Integration by Parts

What is integration by parts?

Integration by parts is a technique used in calculus to integrate products of two functions. When you learned differentiation, you discovered the product rule for finding derivatives of functions that are multiplied together (like $y = x \cdot e^x$ or $y = x \cdot \sin x$ ). Integration by parts is essentially the reverse of this process - it helps us integrate functions that are products of two simpler functions.

The technique becomes necessary because there's no simple "product rule" for integration like there is for differentiation. Instead, we use this special method to break down complex integrals into manageable parts.

chatImportant

Integration by parts is essential when dealing with integrals that cannot be solved using basic integration rules. It's particularly useful when you have products of different types of functions (algebraic, trigonometric, exponential, etc.).

The integration by parts formula

The fundamental formula for integration by parts is:

$\int u \cdot dv = uv - \int v \cdot du$

This formula tells us that to integrate a product, we need to:

Choose one part to differentiate (this becomes $u$ , and its derivative is $du$ )
Choose one part to integrate (this becomes $dv$ , and its integral is $v$ )
Apply the formula to transform our original integral

infoNote

The key to success with integration by parts lies in making the right choice for $u$ and $dv$ . A good choice will simplify the integral, while a poor choice can make it more complex.

The INLATE rule

To decide which function should be $u$ and which should be $dv$ , we use the INLATE rule. This rule gives us a priority order for choosing $u$ :

In = Inverse trigonometric functions (e.g. $\sin^{-1}x$ , $\cos^{-1}x$ )
L = Logarithmic functions (e.g. $\log x$ , $\ln x$ )
A = Algebraic functions (e.g. $3x$ , $x^3$ , polynomials)
T = Trigonometric functions (e.g. $\sin x$ , $\cos x$ , $\tan x$ )
E = Exponential functions (e.g. $e^{4x}$ , $e^{-x}$ )

infoNote

INLATE Memory Tip: The function that appears earlier in this list should be chosen as $u$ (the part we differentiate). The function that appears later should be chosen as $dv$ (the part we integrate).

Setting up the problem

Before applying the formula, we need to split our integral into two separate calculations:

On the left side, we differentiate $u$ to find $du$ . On the right side, we integrate $dv$ to find $v$ . This preparation is essential before substituting into the main formula.

Example 1: $\int xe^{2x}dx$

lightbulbExample

Worked Example: Basic Integration by Parts

Let's work through a straightforward example step by step.

Step 1: Check the INLATE rule ordering

We have $x$ (algebraic) and $e^{2x}$ (exponential)
According to INLATE, algebraic comes before exponential
Therefore: $u = x$ and $dv = e^{2x}dx$

Step 2: Find $du$ and $v$ separately

Differentiating $u = x$ gives us $du = dx$
Integrating $dv = e^{2x}dx$ gives us $v = \frac{1}{2}e^{2x}$

Step 3: Apply the integration by parts formula $\int u \cdot dv = uv - \int v \cdot du$

Substituting our values: $\int x \cdot e^{2x}dx = x \cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x} \cdot dx$

Step 4: Simplify and complete $= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + c$

We can check this answer by differentiating it to see if we get back to our original function.

Example 2: $\int e^x \cos x dx$ (multiple iterations)

Sometimes integration by parts needs to be applied more than once. This example demonstrates that process.

lightbulbExample

Worked Example: Multiple Applications of Integration by Parts

First iteration:

According to INLATE: $\cos x$ (trig) comes before $e^x$ (exponential)
However, the functions are in the wrong order, so we rearrange to: $\int \cos x \cdot e^x dx$
Choose $u = \cos x$ and $dv = e^x dx$
This gives us $du = -\sin x dx$ and $v = e^x$

Applying the formula: $\int \cos x \cdot e^x dx = (\cos x)(e^x) - \int e^x \cdot (-\sin x) dx$ $= (\cos x)(e^x) + \int e^x \sin x dx$

Second iteration: Notice we now have a new integral $\int e^x \sin x dx$ that also requires integration by parts.

Choose $u = \sin x$ and $dv = e^x dx$
This gives us $du = \cos x dx$ and $v = e^x$

Applying the formula again: $\int e^x \sin x dx = (\sin x)(e^x) - \int e^x \cdot \cos x dx$

Clever finishing step: The new integral $\int e^x \cdot \cos x dx$ is exactly the same as what we started with! We can use this to our advantage:

Let $A = \int \cos x \cdot e^x dx$

Then: $A = (\cos x)(e^x) + (\sin x)(e^x) - A$

Solving for $A$ : $2A = e^x(\cos x + \sin x)$

Therefore: $A = \frac{1}{2}e^x(\cos x + \sin x) + C$

chatImportant

When integration by parts leads you back to the original integral, don't give up! This creates an equation you can solve algebraically, as shown in the example above.

When integration by parts doesn't work

Not all integrals can be solved using integration by parts. For composite functions (like functions within functions), we often need different techniques such as substitution. The key is recognising when integration by parts is the appropriate method - typically when you have a clear product of two different types of functions.

chatImportant

Recognition is Key: Integration by parts works best when you have products of different function types (algebraic × exponential, algebraic × trigonometric, etc.). If you're dealing with composite functions or single complex functions, consider other techniques first.

Remember!

bookmarkSummary

Key Points to Remember:

Integration by parts is the reverse of the product rule and uses the formula $\int u \cdot dv = uv - \int v \cdot du$
Use the INLATE rule to decide which function should be $u$ (differentiate) and which should be $dv$ (integrate)
Some problems require multiple applications of integration by parts before reaching a solution
Always check your work by differentiating your final answer to see if you get back to the original integrand
If integration by parts creates increasingly complex integrals, consider whether a different technique might be more appropriate

Integration by Parts (Leaving Cert Applied Maths): Revision Notes