Why this mattersPart 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. ✓
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.
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.
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.
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.
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 outNever 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)
∫ 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|.