Core Integration Techniques

Integration Techniques: The Art of Reverse Engineering 🔄

The Big Picture

Imagine you’re a detective. Someone gives you a finished cake and asks: “What ingredients went into this?”

That’s exactly what integration is! If differentiation is like baking (combining ingredients to make a cake), integration is like un-baking — figuring out what went in.

But here’s the catch: some cakes are tricky. You can’t just look at them and know the recipe. You need special techniques.

Today, we’ll learn two powerful detective tools:

1. Integration by Substitution — The “disguise remover”
2. Integration by Parts — The “divide and conquer” method

🎭 Integration by Substitution

The Story

Imagine you’re trying to read a secret message, but it’s written in a code. The letter “A” is written as “🍎”, “B” as “🍌”, and so on.

What do you do? You substitute the emojis with real letters!

Integration by substitution works the same way. Sometimes integrals look scary because they’re “in disguise.” We substitute a messy part with a simple letter (usually u), solve the easier problem, then switch back.

The Recipe

Step 1: Spot the "inside function" — call it u
Step 2: Find du (the derivative of u)
Step 3: Rewrite everything in terms of u
Step 4: Integrate the simpler expression
Step 5: Substitute back the original variable

Example 1: A Simple Disguise

Problem: Find ∫ 2x(x² + 1)⁵ dx

Think: That (x² + 1) looks messy. Let’s call it u.

Let u = x² + 1
Then du/dx = 2x
So du = 2x dx  ← Look! We have 2x dx in our problem!

Transform: ∫ 2x(x² + 1)⁵ dx = ∫ u⁵ du

Solve the easy version: ∫ u⁵ du = u⁶/6 + C

Switch back: = (x² + 1)⁶/6 + C ✅

Example 2: Trig in Disguise

Problem: Find ∫ sin(x) · cos(x) dx

Two ways to think about this:

Way 1: Let u = sin(x), then du = cos(x) dx ∫ u du = u²/2 + C = sin²(x)/2 + C

Way 2: Let u = cos(x), then du = -sin(x) dx ∫ -u du = -u²/2 + C = -cos²(x)/2 + C

Both are correct! (They differ by a constant — try checking!)

The Golden Rule 🌟

Look for a function AND its derivative sitting together.

If you see f(g(x)) · g’(x), let u = g(x).

🤝 Integration by Parts

The Story

Imagine two friends, U and V, need to carry a heavy box up stairs. Neither can do it alone. So they take turns — U pushes while V rests, then V pushes while U rests.

Integration by parts works like this! When you have two different types of functions multiplied together, you let them take turns being “integrated” and “differentiated.”

The Magic Formula

∫ u dv = uv - ∫ v du

Or in words:

“The integral of U times DV equals U times V minus the integral of V times DU.”

How to Pick U and V?

Use the LIATE rule — pick u in this order of priority:

The one higher in the list becomes u. The other becomes dv.

Example 1: Classic Case

Problem: Find ∫ x · eˣ dx

Pick u and dv using LIATE:

• x is Algebraic (A)
• eˣ is Exponential (E)
• A comes before E, so u = x

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

Apply the formula: ∫ x · eˣ dx = x · eˣ - ∫ eˣ dx = x · eˣ - eˣ + C = eˣ(x - 1) + C ✅

Example 2: The Sneaky Logarithm

Problem: Find ∫ ln(x) dx

Wait — there’s only one function! But we can write it as: ∫ ln(x) · 1 dx

Pick using LIATE:

• ln(x) is Logarithm (L) — highest priority!
• 1 is Algebraic (A)

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

Apply the formula: ∫ ln(x) dx = x · ln(x) - ∫ x · (1/x) dx = x · ln(x) - ∫ 1 dx = x · ln(x) - x + C = x(ln(x) - 1) + C ✅

Example 3: Double Parts (The Boomerang)

Problem: Find ∫ eˣ · sin(x) dx

This one needs integration by parts twice!

Round 1:

u = sin(x)  →  du = cos(x) dx
dv = eˣ dx  →  v = eˣ

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

Round 2: (for the remaining 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

Combine: Let I = ∫ eˣ sin(x) dx

I = eˣ sin(x) - [eˣ cos(x) + I] I = eˣ sin(x) - eˣ cos(x) - I 2I = eˣ(sin(x) - cos(x)) I = eˣ(sin(x) - cos(x))/2 + C ✅

🔄 When to Use Which?

graph TD
    A["Start: Look at the integral"] --> B{Is there a function<br>AND its derivative?}
    B -->|Yes| C["Use SUBSTITUTION"]
    B -->|No| D{Two different<br>function types<br>multiplied?}
    D -->|Yes| E["Use PARTS&lt;br&gt;Apply LIATE rule"]
    D -->|No| F["Try other methods&lt;br&gt;or simplify first"]

Quick Decision Guide

🎯 Practice Mindset

For Substitution:

1. Hunt for the inside function
2. Check if its derivative is nearby
3. If stuck, try u = the messiest part

For Parts:

1. Use LIATE to pick u
2. After one round, is the new integral simpler?
3. If you get the original integral back, use algebra!

💡 Key Takeaways

Substitution is like translation:

Change complicated expressions into simple ones, solve, change back.

Integration by Parts is like teamwork:

When two functions are stuck together, let them take turns helping each other.

Remember:

• Substitution: Look for function + derivative
• Parts: Use LIATE to choose u
• Both: Always add + C at the end!

🚀 You’ve Got This!

Integration might seem hard, but you just learned the two most powerful tools in your toolkit:

1. Substitution — for disguised integrals
2. Parts — for products of different function types

With practice, you’ll start seeing which method to use instantly. It’s like learning to ride a bike — wobbly at first, then second nature.

Now go practice, and watch those integrals bow to your skills! 🎉