The Main Idea 

Think of integration by parts as making a strategic trade. When you’re faced with an integral like [latex]\int x\sin x  dx[/latex], you can’t use substitution because there’s no clear “inside function” and its derivative. Instead, you’re going to swap this integral for a different one that’s easier to handle.

[latex]\int u , dv = uv - \int v  du[/latex]

This formula is your trading mechanism. You’re essentially saying: “I’ll take this product integral and exchange it for something I can actually solve.”

Strategy Tips:

• Choose wisely: Your [latex]u[/latex] should get simpler when differentiated, and your [latex]dv[/latex] should be something you can actually integrate
• Don’t add constants: When finding [latex]v[/latex], skip the “+C” – you’ll add it at the very end
• Check your work: If your new integral [latex]\int v , du[/latex] looks worse than what you started with, try switching your [latex]u[/latex] and [latex]dv[/latex] choices

The Main Idea 

Choosing which part should be [latex]u[/latex] and which should be [latex]dv[/latex] doesn’t have to feel like guesswork. Think of LIATE as your GPS for navigation—it won’t work for every single route, but it’ll get you to your destination most of the time.

The LIATE Priority System:

• Logarithmic functions (like [latex]\ln x[/latex])
• Inverse trig functions (like [latex]\arcsin x[/latex], [latex]\arctan x[/latex])
• Algebraic functions (like [latex]x^2[/latex], [latex]3x[/latex])
• Trigonometric functions (like [latex]\sin x[/latex], [latex]\cos x[/latex])
• Exponential functions (like [latex]e^x[/latex])

The Rule: Pick [latex]u[/latex] to be whichever function type appears first in this list.

LIATE isn’t arbitrary—it’s based on what’s practical. Functions at the top of the list (L and I) are terrible to integrate but get simpler when you differentiate them. Functions at the bottom (T and E) are integration-friendly and stay manageable when differentiated. Algebraic functions (A) sit in the middle because they usually play nice either way.

Strategy Tips:

• Sometimes you need to be creative: If your integral has only one function (like [latex]\int \ln x , dx[/latex]), use the constant function 1 as your second part
• Watch for special cases: Sometimes breaking the LIATE rule makes sense (like choosing [latex]dv = te^{t^2}dt[/latex] instead of just [latex]e^{t^2}dt[/latex] when you can actually integrate the first one)
• Before committing to your choices, ask yourself these two critical questions:
  • “Will my [latex]u[/latex] get simpler when I differentiate it?”
    You want the derivative [latex]du[/latex] to be less complicated than your original [latex]u[/latex]. For example, [latex]\ln x[/latex] becomes [latex]\frac{1}{x}dx[/latex] (simpler), while [latex]e^{x^2}[/latex] becomes [latex]2xe^{x^2}dx[/latex] (more complicated).
  • “Can I actually integrate my [latex]dv[/latex]?”
    There’s no point choosing [latex]dv = e^{x^2}dx[/latex] if you can’t find [latex]v[/latex]. Make sure you can handle the integration step before moving forward.

The Main Idea 

Working with definite integrals using integration by parts is like following the same recipe you already know, but with one additional finishing touch at the end. You still use the exact same formula and decision-making process—you just need to evaluate your result at the upper and lower limits.

[latex]\int_a^b u , dv = uv\big|_a^b - \int_a^b v , du[/latex]

Notice that [latex]uv\big|_a^b[/latex] means you evaluate [latex]uv[/latex] at the upper limit minus [latex]uv[/latex] at the lower limit, just like any other definite integral.

Problem Solving Strategies:

1. Choose [latex]u[/latex] and [latex]dv[/latex] using LIATE or the two-question test
2. Find [latex]du[/latex] and [latex]v[/latex] exactly as before
3. Apply the formula with the limits intact
4. Evaluate [latex]uv\big|_a^b[/latex] at both limits
5. Solve the remaining integral [latex]\int_a^b v , du[/latex] (which might need substitution or other techniques)

You get a specific numerical answer instead of a family of functions plus C. No constant of integration needed!

Always step back and ask if your numerical answer makes sense. For area problems, is your result positive and reasonable compared to basic geometric shapes? For volume problems, does it seem plausible compared to simple cylinders or cones with similar dimensions?