Stewart Calculus · 8th Edition
Chapter 7: Techniques of Integration
Every key formula from Stewart Calculus Chapter 7, in one searchable page. Click a card to study it — worked examples included.
7.1 · INTEGRATION BY PARTS
∫u dv = uv − ∫v du
Pick u by LIATE: Log, Inverse trig, Algebraic, Trig, Exponential.
WORKED EXAMPLE
∫x ex dx: u = x, dv = exdx ⇒
x ex − ∫ex dx = (x − 1)ex + C.
x ex − ∫ex dx = (x − 1)ex + C.
WORKED EXAMPLE
∫ex sin x dx: parts twice, solve for I ⇒
ex(sin x − cos x)/2 + C.
ex(sin x − cos x)/2 + C.
∫sec3x dx = ½sec x tan x + ½ln|sec x + tan x| + C
∫xn ln x dx: u = ln x (LIATE: log beats algebraic).
7.2 · TRIG INTEGRALS
∫sinmx cosnx dx
n odd: save one cos, u = sin x.
m odd: save one sin, u = cos x.
Both even: half-angle formulas.
m odd: save one sin, u = cos x.
Both even: half-angle formulas.
∫tanmx secnx dx
n even: save sec², u = tan x.
m odd: save sec·tan, u = sec x.
m odd: save sec·tan, u = sec x.
Half-angle
sin²x = (1 − cos 2x)/2, cos²x = (1 + cos 2x)/2
cot / csc
Mirror the tan/sec rules with cot and csc.
WORKED EXAMPLE
∫sin³x cos²x dx = ∫(1−u²)u² du, u = cos x ⇒
−cos³x/3 + cos5x/5 + C.
−cos³x/3 + cos5x/5 + C.
Both exponents even? Half-angle first, then reduce.
7.3 · TRIG SUBSTITUTION
√(a² − x²) ⇒ x = a sin θ
√(a² + x²) ⇒ x = a tan θ
√(x² − a²) ⇒ x = a sec θ
√(a² + x²) ⇒ x = a tan θ
√(x² − a²) ⇒ x = a sec θ
KEY NOTES
- Draw the right triangle — then back-substitute to x.
WORKED EXAMPLE
∫ dx/√(9 − x²): x = 3 sin θ ⇒ ∫ dθ = arcsin(x/3) + C.
∫dx/√x²+1 = ln|x + √x²+1| + C
For x = a sec θ: dx = a sec θ tan θ dθ.
WORKED EXAMPLE
∫x²/√(4 − x²) dx: x = 2 sin θ ⇒ ∫4 sin²θ dθ
= 2θ − sin 2θ + C ⇒ triangle back-sub.
= 2θ − sin 2θ + C ⇒ triangle back-sub.
7.4 · PARTIAL FRACTIONS
Setup
Long-divide first if deg(num) ≥ deg(den).
Ax − a + Bx − b · Ax − a + B(x − a)² · Ax + Bquadratic
WORKED EXAMPLE
∫(2x+3)/(x²+x) dx:
= ∫[3/x − 1/(x+1)] dx = 3 ln|x| − ln|x+1| + C.
= ∫[3/x − 1/(x+1)] dx = 3 ln|x| − ln|x+1| + C.
Repeated: Ax−1 + B(x−1)²
Irreducible quadratic ⇒ complete the square ⇒ arctan form.
7.7 · APPROXIMATE INTEGRATION
Trapezoidal
Tn = Δx2[f(x₀) + 2f(x₁) + … + f(xn)]
Simpson’s (n even)
Sn = Δx3[f(x₀) + 4f(x₁) + 2f(x₂) + … + f(xn)]
Errors: |ET| ≤ K(b−a)³/(12n²), |ES| ≤ K(b−a)5/(180n4).
Midpoint
Mn = Δx[f(x̄1) + … + f(x̄n)]; |EM| ≤ K(b−a)³/(24n²)
Simpson needs n even. Bigger n ⇒ smaller error.
WORKED EXAMPLE
Simpson n = 2 for ∫01 x² dx:
S₂ = (0.5/3)[0 + 4(0.25) + 1] = 1/3 (exact!).
S₂ = (0.5/3)[0 + 4(0.25) + 1] = 1/3 (exact!).
Type 1: infinite limits
∫a∞ f(x) dx = limt→∞ ∫at f(x) dx
Type 2: discontinuous integrand
Split the integral at every interior bad point.
p-test
∫1∞ dx/xp converges ⟺ p > 1
WORKED EXAMPLE
∫1∞ dx/x² = [−1/x]1∞ = 1 (converges).
Comparison
If 0 ≤ f ≤ g and ∫g converges ⇒ ∫f converges.
WORKED EXAMPLE
∫1∞ dx/(x²+1) ≤ ∫1∞ dx/x² = 1 ⇒ converges.
WATCH OUT!
Mistakes that cost points
- +C on every antiderivative
- Partial fractions: long-divide first if improper
- Trig sub: convert back to x (use the triangle!)
- Improper: split at every interior bad point
- Simpson’s rule needs n EVEN
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