# Chapter 19: Integration Techniques, Part Two

> When substitution and parts run out of road: trig identities (sin²x, cos²x), parity rules for powers of sin/cos/tan/sec, the three trig substitutions for square roots (sine, tangent, secant), and completing the square — with a summary table.

**Canonical:** https://calculus.kitzo.work/chapter/integration-2  
**Level:** Advanced  
**Estimated time:** 55 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Trig identities in integrals
- Powers of sin and cos
- Powers of tan and sec
- Trig substitution type 1
- Trig substitution type 2 & 3
- Completing the square

**Why this matters**

Part One gave you substitution, parts and fractions. But some integrands wear armor made of trig functions and square roots — and only identities and geometry can pierce it. This chapter is that toolkit.

## 1 · Trig identities as antiderivative super-glue

Three identities do the heavy lifting:

  - sin²x = (1 − cos 2x)/2 ·  cos²x = (1 + cos 2x)/2

  - sin²x + cos²x = 1  ·  1 + tan²x = sec²x

  - tan²x = sec²x − 1  ·  cos²x = 1 − sin²x

The half-angle pair is what turns ∫ sin²x dx from a puzzle into routine algebra.

**Worked example**

**Example 1 — ∫ sin²x dx.**

Step 1: replace sin²x with (1 − cos 2x)/2.

Step 2: ∫ (1 − cos 2x)/2 dx = x/2 − sin 2x/4.

Step 3: answer: **x/2 − sin(2x)/4 + C**. Check by differentiating: ½ − ½cos 2x = sin²x. ✓

## 2 · Powers of sin and cos

The strategy depends on parity:

  - **Odd power anywhere:** peel off one factor, convert the rest using 1 − cos²x = sin²x (or 1 − sin²x = cos²x), then substitute u = cos x (or u = sin x).

  - **Both powers even:** fall back on the half-angle identities — sin²x cos²x = (1/8)(1 − cos 4x) handles the hardest case.

**Worked example**

**Example 2 — ∫ cos³x dx.**

Step 1: cos³x = cos²x·cos x = (1 − sin²x)cos x.

Step 2: u = sin x, du = cos x dx.

Step 3: ∫ (1 − u²)du = u − u³/3.

Step 4: answer: **sin x − sin³x/3 + C**.

## 3 · Powers of tan and sec

Two reflexes cover nearly every tan/sec integral:

  - See a **sec²x** factor? Save it as du with u = tan x, and convert the remaining tan²x using tan²x = sec²x − 1.

  - See a **tan x·sec x** pair? Save it with u = sec x.

Mantra: keep one sec²x *or* one (sec x tan x), whichever the problem offers, and feed the rest to an identity.

**Worked example**

**Example 3 — ∫ tan³x dx.**

Step 1: tan³x = tan x·(sec²x − 1).

Step 2: u = tan x, du = sec²x dx handles the first piece: ∫ tan x sec²x dx = tan²x/2.

Step 3: the leftover is ∫ tan x dx = −ln|cos x| (since d/dx cos x = −sin x).

Step 4: answer: **tan²x/2 + ln|cos x| + C**.

Also memorize ∫ sec x dx = ln|sec x + tan x| + C and ∫ sec x tan x dx = sec x + C.

Pure powers of sec need a reduction formula (a parts argument trading secⁿ for secⁿ⁻²) — know it exists; reach for it last.

## 4 · Type 1: √(a² − x²)

Use `x = a sin θ`, θ ∈ [−π/2, π/2]. Then a² − x² = a²cos²θ and the root collapses to a cos θ. Back-substitute with a θ-triangle.

**Worked example**

**Example 4 — ∫ √(9 − x²) dx.**

Step 1: a = 3. Put x = 3 sin θ, dx = 3 cos θ dθ.

Step 2: √(9 − 9sin²θ) = 3cos θ (cos θ ≥ 0 on our interval).

Step 3: ∫ 9cos²θ dθ = 9·(θ/2 + sin 2θ/4) + C = (9/2)θ + (9/4)sin 2θ + C.

Step 4: sin 2θ = 2sin θcos θ, so this is (9/2)θ + (9/2)sin θcos θ.

Step 5: arcsin(x/3) back for θ, sin θ = x/3, cos θ = √(9−x²)/3:

answer: **(x/2)·√(9 − x²) + (9/2)·arcsin(x/3) + C**.

## 5 · Types 2 and 3: √(a² + x²), √(x² − a²)

**Type 2:** √(a² + x²) → `x = a tan θ`. Then a² + x² = a²sec²θ.

**Type 3:** √(x² − a²) → `x = a sec θ`. Then x² − a² = a²tan²θ.

**Worked example**

**Example 5 — ∫ dx/√(x² + 9).**

Step 1: a = 3. Put x = 3 tan θ, dx = 3 sec²θ dθ.

Step 2: √(x² + 9) = 3 sec θ.

Step 3: ∫ 3 sec²θ dθ / (3 sec θ) = ∫ sec θ dθ = ln|sec θ + tan θ| + C.

Step 4: tan θ = x/3, sec θ = √(x²+9)/3 — the /3 absorbs into C:

answer: **ln|x + √(x² + 9)| + C**.

## 6 · Complete the square first, then the table

Quadratics in disguise must be completed first: x² + 4x + 5 = (x + 2)² + 1 — Type 2 with u = x + 2.

**Summary table:**

  - √(a² − x²) → x = a sin θ → a cos θ

  - √(a² + x²) → x = a tan θ → a sec θ

  - √(x² − a²) → x = a sec θ → a tan θ

  - ax² + bx + c → complete the square first, then pattern-match

**Watch out**

Never guess the sign of cos θ after back-substituting — draw the θ-triangle. And completing the square is mandatory whenever b ≠ 0: x = a sec θ on an uncompleted square goes nowhere.

Chapter formula sheet
sin²x = (1 − cos 2x)/2  ·  cos²x = (1 + cos 2x)/2  ·  tan²x = sec²x − 1

∫ sin²x dx = x/2 − sin 2x/4 + C  ·  ∫ cos²x dx = x/2 + sin 2x/4 + C

Odd power: save one factor → u = the other trig function

∫ tan²x dx = tan x − x + C  ·  ∫ sec²x dx = tan x + C  ·  ∫ sec x dx = ln|sec x + tan x| + C

√(a² − x²) → x = a sin θ  ·  √(a² + x²) → x = a tan θ  ·  √(x² − a²) → x = a sec θ

∫ dx/(x² + a²) = (1/a) arctan(x/a) + C  ·  complete the square: ax² + bx + c → a(x + b/2a)² + (c − b²/4a)

## Common mistakes

  - Attacking ∫ sin⁵x dx with half-angle identities — odd powers always yield to u = cos x.

  - Choosing x = a sec θ for √(a² − x²) — the minus sign demands sine.

  - Back-substituting θ but leaving sin θ, cos θ in the final answer.

  - Forgetting the factor a from dx = a cos θ dθ (or a sec²θ dθ).

  - Trying trig substitution on x² + bx + c without completing the square first.

  - Dropping the absolute value in ln|sec x + tan x|.

## Practice exercises

1. ∫ sin²x dx equals:

- A. x/2 − sin(2x)/4 + C
- B. x/2 + sin(2x)/4 + C
- C. −cos²x/2 + C
- D. sin x/2 + C

2. Evaluate ∫₀ π/2 cos x dx.

3. Evaluate ∫ tan²x dx. (Include + C.)

4. Best first move for ∫ sin⁵x dx:

- A. apply the half-angle identity twice
- B. peel off one sin x, write sin⁴x = (1 − cos²x)², substitute u = cos x
- C. use integration by parts with u = sin x
- D. write sin⁵x = sin x/(cos²x)

5. Evaluate ∫₀ π/2 sin x·cos x dx.

6. For ∫ √(25 − x²) dx, write the trig substitution (x = …).

7. Completing the square, x² + 6x + 10 equals:

- A. (x + 3)² + 1
- B. (x + 6)² + 10
- C. (x + 3)² − 1
- D. (x − 3)² + 1

8. Evaluate ∫ dx/((x + 1)² + 1). (Include + C.)

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