Integration By Parts Examples With Solutions

Integration by Parts Examples with Solutions

Ever stare at an integral that just refuses to simplify? Day to day, you've tried substitution, you've tried algebraic manipulation, and nothing's working. That's usually the moment integration by parts enters the conversation. Still, it's one of those techniques that looks intimidating at first but becomes second nature once you've worked through a handful of examples. The trick is understanding when* to use it and how to choose which part of the expression plays which role. Let's walk through it together.

What Is Integration by Parts

Integration by parts is a method derived from the product rule of differentiation. When you have an integral of a product of two functions — something like ∫ u dv — and neither function alone is easy to integrate on its own, this technique lets you rewrite the integral in a form that's more manageable.

The formula looks like this:

∫ u dv = uv − ∫ v du

That's it. That's the whole engine. But the real skill lies in choosing u and dv wisely, because a bad choice can make the new integral harder than the original one.

The Core Idea in Plain Language

Think of it this way. You're taking two things that are multiplied together inside an integral, and you're peeling them apart so that one part gets differentiated (becomes simpler) and the other part gets integrated. The goal is to trade a hard integral for an easier one.

Why It Matters

You'll encounter integration by parts in nearly every course that covers calculus beyond the basics. It shows up in physics when calculating things like moments of inertia, in probability when working with expected values of continuous distributions, and in engineering when solving differential equations.

Without this technique, a large class of integrals — especially those involving logarithms, inverse trigonometric functions, and products of polynomials with exponentials or trig functions — would be essentially inaccessible using elementary methods.

How Integration by Parts Works

The LIATE Rule for Choosing u

One of the most practical heuristics out there is the LIATE rule. It gives you a priority order for deciding which factor in a product should become u:

• L — Logarithmic functions (ln x, log₂ x)
• I — Inverse trigonometric functions (arctan x, arcsin x)
• A — Algebraic functions (polynomials like x², x³)
• T — Trigonometric functions (sin x, cos x)
• E — Exponential functions (eˣ, 2ˣ)

The function higher on the list gets chosen as u, and the one lower on the list becomes dv. The reasoning is that functions higher on LIATE tend to simplify when differentiated, while those lower tend to be easy to integrate.

That said, LIATE is a guideline, not a law. Sometimes you need to break it — and you'll learn when that happens by doing the work.

Step-by-Step Process

Here's the general workflow:

1. Identify the product of two functions in the integrand.
2. Choose u and dv based on LIATE (or intuition, if you've practiced enough).
3. Compute du (by differentiating u) and v (by integrating dv).
4. Plug everything into the formula: uv − ∫ v du.
5. Evaluate the new integral. If it's still complicated, you may need to apply integration by parts again.

Integration by Parts Examples with Solutions

Example 1: A Polynomial Times an Exponential

Problem: ∫ x eˣ dx

This is a classic first example because it's clean and the method is straightforward.

Following LIATE, the algebraic function x goes above the exponential eˣ, so:

• u = x → du = dx
• dv = eˣ dx → v = eˣ

Applying the formula:

∫ x eˣ dx = x eˣ − ∫ eˣ dx

The remaining integral is trivial:

= x eˣ − eˣ + C

You can factor this if you like:

= eˣ (x − 1) + C

That's the full solution. One round of integration by parts did the job.

Example 2: A Logarithmic Function

Problem: ∫ ln x dx

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This one might look odd because there's no obvious "product." But you can think of it as ∫ 1 · ln x dx, which gives you two factors to work with.

Using LIATE, the logarithmic function gets priority as u:

• u = ln x → du = (1/x) dx
• dv = dx → v = x

Plugging into the formula:

∫ ln x dx = x ln x − ∫ x · (1/x) dx

= x ln x − ∫ 1 dx

= x ln x − x + C

Example 3: A Polynomial Times a Trigonometric Function

Problem: ∫ x sin x dx

Again, the algebraic function x is higher on LIATE than the trigonometric function sin x:

• u = x → du = dx
• dv = sin x dx → v = −cos x

Applying the formula:

∫ x sin x dx = −x cos x − ∫ (−cos x) dx

= −x cos x + ∫ cos x dx

= −x cos x + sin x + C

Example 4: Repeated Application — The Cyclical Case

Problem: ∫ eˣ sin x dx

This is where things get interesting. You'll need to apply integration by parts twice, and something almost magical happens.

First application:

• u = sin x → du = cos x dx
• dv = eˣ dx → v = eˣ

∫ eˣ sin x dx = eˣ sin x − ∫ eˣ cos x dx

Now apply integration by parts to the new integral:

• u = cos x → du = −sin x dx
• dv = eˣ dx → v = eˣ

∫ eˣ cos x dx = eˣ cos x − ∫ eˣ (−sin x) dx

= eˣ cos x + ∫ eˣ sin x dx

Now substitute this back into the first equation:

∫ eˣ sin x dx = eˣ sin x − (eˣ cos x + ∫ eˣ sin x dx)

= eˣ sin x − eˣ cos x − ∫ eˣ sin x dx

Here's the trick: the original integral shows up on both sides of the equation. Add ∫ eˣ sin x dx to both sides:

2 ∫ eˣ sin x dx = eˣ sin x − eˣ cos x

Divide by 2:

∫ eˣ sin x dx = (eˣ / 2)(sin x − cos x) + C

This cyclical pattern comes up often with products of exponentials and

trigonometric functions.

Summary and Tips for Success

Integration by parts is a powerful tool that transforms a difficult integral into a simpler one by shifting the "burden" from a function that is hard to integrate to one that is easy to differentiate. Even so, it is not always a "one-and-done" method. To master this technique, keep these final strategies in mind:

• The LIATE Rule is Your Compass: Always use the Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential hierarchy to choose your $u$. Choosing the wrong $u$ can often make the integral more complicated rather than simpler.
• Watch Your Signs: The most common error in integration by parts isn't the calculus itself, but the algebra. Forgetting to distribute a negative sign when substituting the second integral (as seen in Example 4) will lead to an incorrect result.
• Recognize the Patterns: If you see an integral that returns to its original form after two rounds of integration by parts, don't panic—you haven't made a mistake. You have simply encountered a "cyclical" integral. Just use algebra to solve for the integral as if it were a variable.
• Don't Forget "+ C": Since integration is the reverse of differentiation, always remember to include the constant of integration for indefinite integrals.

Conclusion

Integration by parts is more than just a formula; it is a strategic approach to breaking down complex mathematical structures. Think about it: while it may initially seem daunting to deal with multiple layers of integration or the "looping" integrals found in exponential-trigonometric products, these patterns become much clearer with consistent practice. Once you become comfortable identifying your $u$ and $dv$ and managing the resulting algebraic steps, you will find that integration by parts is one of the most versatile and essential tools in your calculus toolkit.