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U-substitution undoes the chain rule. Spot the pattern f(g(x))·g'(x): the inner function and its derivative both appear (up to a constant). Set u = inner, du = inner' dx, replace, integrate, back-substitute.
∫ f(g(x)) · g'(x) dx = ∫ f(u) du, u = g(x), du = g'(x) dx| Integrand has… | Try u = | Result |
|---|---|---|
| (stuff)ⁿ · stuff' | stuff | u^(n+1)/(n+1) |
| e^(stuff) · stuff' | stuff | e^u |
| (stuff'/stuff) | stuff | ln|u| |
| sin(ax) or cos(ax) | ax | −cos u/a or sin u/a |
Inverse trig pattern: ∫ dx/√(a²−x²) = arcsin(x/a) + C, ∫ dx/(a²+x²) = (1/a) arctan(x/a) + C. Memorize these — they're free points.
Indefinite integrals always need +C. After u-sub on a definite integral, either change limits to u-values or back-sub before evaluating. Mixing the two is the most common 1-point loss on Calc 2.
| Form | Sub | Identity | dx |
|---|---|---|---|
| √(a²−x²) | x = a sin θ | 1−sin² = cos² | a cos θ dθ |
| √(a²+x²) | x = a tan θ | 1+tan² = sec² | a sec²θ dθ |
| √(x²−a²) | x = a sec θ | sec²−1 = tan² | a sec θ tan θ dθ |
After integrating in θ, draw a reference triangle to back-substitute to x. The triangle's sides come from the substitution.
∫ dx/√(9 − x²) =[x=3 sinθ]= ∫ dθ = arcsin(x/3) + CUse when integrand is a proper rational function P(x)/Q(x) with deg P < deg Q. If improper, do polynomial long division first.
∫ 1/(x²−1) dx = ½ ∫ [1/(x−1) − 1/(x+1)] dx = ½ ln|(x−1)/(x+1)| + CIf deg numerator ≥ deg denominator, you must long-divide first. Skipping this gives nonsense partial fractions and a wrong answer with full work shown — graders will deduct heavily.
∫ u dv = u·v − ∫ v duPick u = something that gets simpler when differentiated. Pick dv = something you can integrate. The remaining integral ∫ v du should be easier than the original.
| Letter | Class | Example |
|---|---|---|
| L | Logarithm | ln x |
| I | Inverse trig | arctan x |
| A | Algebraic | x², x³ |
| T | Trig | sin x, cos x |
| E | Exponential | eˣ |
∫ ln x dx: u = ln x, dv = dx → ln x · x − ∫ 1 dx = x ln x − x + CReduction formulas: ∫ xⁿ eˣ dx = xⁿ eˣ − n ∫ x^(n−1) eˣ dx — apply IBP n times until it bottoms out.
If du gets uglier than u, you picked wrong. Switch immediately. Most wasted exam minutes on IBP come from stubbornly pushing through a bad u choice. LIATE saves you.
f(x) = Σn=0∞ f⁽ⁿ⁾(a) · (x − a)ⁿ / n!A Taylor series approximates f(x) near x = a using its derivatives at a. Maclaurin = Taylor at a = 0. Each higher-order term improves accuracy near a; far from a, you need more terms.
| Function | Series | Converges |
|---|---|---|
| eˣ | 1 + x + x²/2! + x³/3! + … | all x |
| sin x | x − x³/3! + x⁵/5! − … | all x |
| cos x | 1 − x²/2! + x⁴/4! − … | all x |
| 1/(1−x) | 1 + x + x² + x³ + … | |x| < 1 |
| ln(1+x) | x − x²/2 + x³/3 − … | −1 < x ≤ 1 |
Trick: for ∫ sin(x²) dx (no closed form), substitute into the sin series, integrate term-by-term — that's how you get an answer.
It's n! in the denominator, not n. Off-by-one factorials kill problems. Also: don't forget to raise (x − a) to n, not just x.
Given x = f(t), y = g(t), the curve is traced as t increases. Use t as the natural progress parameter.
dy/dx = (dy/dt) / (dx/dt), d²y/dx² = (d/dt)[dy/dx] / (dx/dt)Arc length: L = ∫ab √((dx/dt)² + (dy/dt)²) dt| Convert | Formula |
|---|---|
| polar → Cartesian | x = r cos θ, y = r sin θ |
| Cartesian → polar | r = √(x² + y²), θ = arctan(y/x) |
| r = constant | circle of radius r at origin |
| θ = constant | ray from origin |
A = ½ ∫αβ r(θ)² dθ — area swept by ray from θ=α to θ=β. Note the square and 1/2.Polar arc length: L = ∫ √(r² + (dr/dθ)²) dθCommon polar curves: r = a (circle), r = a + b cos θ (cardioid/limaçon), r = a sin(nθ) (rose with n or 2n petals).
Polar area uses ½ ∫ r² dθ, not ∫ y dx. Forgetting the 1/2 or the square loses the entire problem. And use the right θ-bounds — sometimes a curve traces twice as θ goes 0 to 2π.
V = π ∫ab [R(x)]² dxStack disks perpendicular to the axis of rotation. Each disk has radius R(x) — the distance from axis to curve. Use when slicing perpendicular gives clean disks.
V = 2π ∫ab x · h(x) dx (rotate around y-axis)Stack thin cylindrical shells parallel to the axis. Radius = x (distance to axis), height = h(x) = top − bottom. Use when shells are easier to set up than disks.
| Pick disk if… | Pick shell if… |
|---|---|
| slicing ⊥ axis is clean | slicing ∥ axis is clean |
| function is y of x and rotating around x | function is y of x and rotating around y |
| region bounded by curve and axis | region between two curves with hole |
L = ∫√(1 + (dy/dx)²) dx — derivative under root. Almost no closed-form examples are pretty; expect L'Hôpital or Simpson's rule.Rotate around x-axis with f(x): use disk in x. Rotate around y-axis with f(x): use shell in x OR rewrite as x = g(y) and disk in y. Picking the wrong frame turns a 5-min problem into 20.
| Test | Use when | Concl. |
|---|---|---|
| nth term | always first | if an ↛ 0 ⇒ diverges |
| Geometric | an = ar^n | conv ⇔ |r| < 1 |
| p-series | 1/n^p | conv ⇔ p > 1 |
| Integral | an = f(n), f decreasing | matches ∫ f |
| Comparison | vs known series | squeeze logic |
| Limit comparison | messy an/bn | same fate as bn |
| Ratio | factorials, n!, x^n | L < 1 conv, L > 1 div |
| Root | (stuff)^n | L < 1 conv |
| Alternating | (−1)^n an, an ↘ 0 | conv |
Geometric: Σ ar^n = a/(1 − r) for |r| < 1If lim an = 0, the series might converge or diverge — the test is inconclusive. Σ 1/n has an → 0 and still diverges (harmonic). nth term only proves divergence, never convergence.
| Pattern | Use § from | Move |
|---|---|---|
| (stuff)ⁿ · stuff' | § ① | u-sub |
| x · eˣ, xⁿ · sin/cos, ln x | § ② | integration by parts (LIATE) |
| √(a²−x²), √(a²+x²), √(x²−a²) | § ⑥ | trig substitution |
| P(x)/Q(x), proper rational | § ⑥ | partial fractions |
| P(x)/Q(x), improper | § ⑥ | long divide first, THEN PF |
| 'volume rotated about' | § ③ | disk / washer / shell |
| 'arc length of' | § ③ | ∫√(1+(f')²) dx |
| 'does Σ converge?' | § ④ | nth term first, then test ladder |
| factorials, n^n in series | § ④ | ratio or root test |
| (−1)^n in series | § ④ | alternating series test |
| 'radius of convergence' | § ④ | ratio test in x, check endpoints |
| 'approximate f(x)' | § ⑤ | Taylor at nearest a |
| 'Maclaurin of g(x)' | § ⑤ | substitute / differentiate / integrate known series |
| x = f(t), y = g(t) | § ⑦ | parametric: dy/dx = (dy/dt)/(dx/dt) |
| r = f(θ), 'area inside' | § ⑦ | ½ ∫ r² dθ |
If you've spent 5 minutes on a method and the integral isn't simplifying, stop and try a different one. Calc 2 problems are designed for one specific technique — going down the wrong path is the #1 time waster.
Volumes are in cubic units. Polar bounds run over θ, not x. Series partial sums vs full sums differ. Always state what you computed at the end so partial credit isn't lost to ambiguity.