Trigonometric Integrals

🎭 The Dance of Trigonometric Integrals

Imagine you’re a chef combining ingredients. Sometimes you need to mix things in a special way to get the perfect dish. Trigonometric integrals are exactly like that—combining sines and cosines using special “recipes” to solve tricky problems!

🌟 The Big Picture

When we see integrals with sines and cosines multiplied together or raised to powers, we can’t just integrate them directly. We need clever tricks—like a magician pulling a rabbit out of a hat!

Our Two Main Tricks:

1. Powers of Sine and Cosine → Use identities to simplify
2. Trigonometric Substitution → Replace x with a trig function

🎪 Part 1: Powers of Sine and Cosine

What Are We Solving?

Integrals that look like:

$\int \sin^m(x) \cos^n(x) , dx$

Where m and n are numbers (like 2, 3, 4…).

🔑 The Secret: Look at the Powers!

Think of it like sorting laundry:

• Odd power? → Easy to separate!
• Even power? → Use a special folding trick!

Case 1: One Power is ODD 🎯

The Trick: Save one factor, convert the rest!

Example: ∫ sin³(x) cos²(x) dx

Step 1: Sin has odd power (3). Save one sin(x): $= \int \sin^2(x) \cos^2(x) \cdot \sin(x) , dx$

Step 2: Convert sin²(x) using the magic identity: $\sin^2(x) = 1 - \cos^2(x)$

$= \int (1 - \cos^2(x)) \cos^2(x) \cdot \sin(x) , dx$

Step 3: Let u = cos(x), then du = -sin(x) dx: $= -\int (1 - u^2) u^2 , du$ $= -\int (u^2 - u^4) , du$ $= -\frac{u^3}{3} + \frac{u^5}{5} + C$ $= -\frac{\cos^3(x)}{3} + \frac{\cos^5(x)}{5} + C$

Case 2: Both Powers are EVEN 🎨

The Trick: Use half-angle identities!

The Magic Formulas: $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ $\cos^2(x) = \frac{1 + \cos(2x)}{2}$

Example: ∫ sin²(x) dx

Step 1: Apply half-angle identity: $= \int \frac{1 - \cos(2x)}{2} , dx$

Step 2: Split and integrate: $= \frac{1}{2} \int 1 , dx - \frac{1}{2} \int \cos(2x) , dx$ $= \frac{x}{2} - \frac{\sin(2x)}{4} + C$

🎮 Quick Reference Chart

🚀 Part 2: Trigonometric Substitution

The Big Idea 💡

Sometimes an integral has a square root that looks scary:

• √(a² - x²)
• √(a² + x²)
• √(x² - a²)

The Magic: Replace x with a trig function, and the square root disappears!

Think of it like putting on disguises. The ugly square root transforms into something beautiful!

🎭 The Three Disguises

┌─────────────────────────────────────────────┐
│  Expression      │  Substitute  │  Why?    │
├─────────────────────────────────────────────┤
│  √(a² - x²)      │  x = a sin θ │  Uses    │
│                  │              │  1-sin²=cos²│
├─────────────────────────────────────────────┤
│  √(a² + x²)      │  x = a tan θ │  Uses    │
│                  │              │  1+tan²=sec²│
├─────────────────────────────────────────────┤
│  √(x² - a²)      │  x = a sec θ │  Uses    │
│                  │              │  sec²-1=tan²│
└─────────────────────────────────────────────┘

🌈 Substitution 1: √(a² - x²) → Use x = a sin θ

Example: ∫ √(9 - x²) dx

Step 1: Identify: a² = 9, so a = 3

Step 2: Substitute: x = 3 sin θ, so dx = 3 cos θ dθ

Step 3: Transform the square root: $\sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta}$ $= \sqrt{9(1 - \sin^2\theta)}$ $= 3\sqrt{\cos^2\theta} = 3\cos\theta$

Step 4: The integral becomes: $\int 3\cos\theta \cdot 3\cos\theta , d\theta = 9\int \cos^2\theta , d\theta$

Step 5: Use half-angle (from Part 1!): $= 9 \cdot \frac{1}{2}(\theta + \sin\theta\cos\theta) + C$ $= \frac{9\theta}{2} + \frac{9\sin\theta\cos\theta}{2} + C$

Step 6: Convert back to x:

• θ = arcsin(x/3)
• sin θ = x/3
• cos θ = √(9-x²)/3

$= \frac{9}{2}\arcsin\frac{x}{3} + \frac{x\sqrt{9-x^2}}{2} + C$

🌟 Substitution 2: √(a² + x²) → Use x = a tan θ

Example: ∫ 1/√(4 + x²) dx

Step 1: Identify: a² = 4, so a = 2

Step 2: Substitute: x = 2 tan θ, dx = 2 sec² θ dθ

Step 3: Transform: $\sqrt{4 + x^2} = \sqrt{4 + 4\tan^2\theta}$ $= 2\sqrt{1 + \tan^2\theta} = 2\sec\theta$

Step 4: The integral: $\int \frac{2\sec^2\theta}{2\sec\theta} , d\theta = \int \sec\theta , d\theta$ $= \ln|\sec\theta + \tan\theta| + C$

Step 5: Convert back (draw the triangle!):

• tan θ = x/2
• sec θ = √(4+x²)/2

$= \ln\left|\frac{\sqrt{4+x^2}}{2} + \frac{x}{2}\right| + C$ $= \ln\left|\sqrt{4+x^2} + x\right| + C’$

⚡ Substitution 3: √(x² - a²) → Use x = a sec θ

Example: ∫ √(x² - 16)/x dx

Step 1: Identify: a² = 16, so a = 4

Step 2: Substitute: x = 4 sec θ, dx = 4 sec θ tan θ dθ

Step 3: Transform: $\sqrt{x^2 - 16} = \sqrt{16\sec^2\theta - 16}$ $= 4\sqrt{\sec^2\theta - 1} = 4\tan\theta$

Step 4: The integral: $\int \frac{4\tan\theta}{4\sec\theta} \cdot 4\sec\theta\tan\theta , d\theta$ $= 4\int \tan^2\theta , d\theta$ $= 4\int (\sec^2\theta - 1) , d\theta$ $= 4(\tan\theta - \theta) + C$

Step 5: Convert back:

• sec θ = x/4
• tan θ = √(x²-16)/4
• θ = arcsec(x/4)

$= \sqrt{x^2-16} - 4,\text{arcsec}\frac{x}{4} + C$

🎯 The Triangle Trick

When converting back to x, draw a right triangle!

For x = a sin θ:         For x = a tan θ:         For x = a sec θ:

      /|                       /|                       /|
     / |                      / |                      / |
  a /  | x               √(a²+x²)/  | x              x /  |√(x²-a²)
   /   |                    /   |                    /   |
  /θ___|                   /θ___|                   /θ___|
   √(a²-x²)                  a                        a

🌟 Summary: Your Action Plan

graph TD
    A["See a trig integral?"] --> B{What type?}
    B -->|sin^m × cos^n| C{Check powers}
    B -->|Has √...| D{What's inside?}

    C -->|One is odd| E["Save one, convert rest"]
    C -->|Both even| F["Use half-angle formulas"]

    D -->|√a²-x²| G["x = a sin θ"]
    D -->|√a²+x²| H["x = a tan θ"]
    D -->|√x²-a²| I["x = a sec θ"]

    E --> J["U-substitution"]
    G --> K["Draw triangle at end"]
    H --> K
    I --> K

💫 Final Tips

1. For powers: Always ask “Is one power odd?”
2. For square roots: Match the pattern to pick your substitution
3. Don’t forget: After trig sub, convert back to x!
4. Practice makes perfect: These become automatic with repetition

You’ve now got the keys to unlock any trigonometric integral. Go forth and integrate with confidence! 🎉