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Grade 12 / High 3 (age 17-18)

Integration Methods

Integration Methods

Integration feels hard because integrands come in so many shapes, so the trick is to pick the right tool for each one. Substitution rewrites a messy problem in x as a simpler one in a new variable u, while integration by parts trades a hard integral for an easier one. A square root hints at a trig substitution and a fraction hints at partial fractions, so the form itself tells you the strategy. Drag the upper-limit slider to watch the x-world turn into the u-world, and to see how the area under xeˣ builds up.

Substitution
2
🔄 Change Worlds
①Left (x-world): 2x·cos(x²) — complicated
②Substitute u = x² → right (u-world): cos(u) — simple!
③Integrate then return to x: sin(x²) + C
Substitution Formula
∫ f(g(x))·g'(x) dx = ∫ f(u) du (u = g(x))
If g(x) = u then g'(x)dx = du
Integration by Parts
1.5
IBP Formula
∫ u dv = uv − ∫ v du
u: easy to differentiate; dv: easy to integrate
🧩 LIATE Rule
①Choose u in order: Log → Inverse trig → Algebraic → Trig → Exponential
②E.g., ∫xeˣdx → u = x (algebraic), dv = eˣdx
③Result: xeˣ − ∫eˣdx = xeˣ − eˣ + C
Common Substitution Patterns
Trig Substitution
√(a²−x²) → x = a sin θ, √(x²−a²) → x = a sec θ
Pick a trig substitution based on the radical form
Partial Fractions
1(x-a)(x-b) = Ax-a + Bx-b
Decompose a rational function and integrate piece by piece
💡 Strategy
①Composite form → substitution
②Product of functions → integration by parts
③Radicals → trig substitution
④Rational functions → partial fractions
Common Antiderivatives
Exponential / Log
∫ eˣ dx = eˣ + C, ∫ 1x dx = ln|x| + C
Integral of eˣ is itself; integral of 1/x is ln|x|
Trig
∫ sin x dx = −cos x + C, ∫ cos x dx = sin x + C
Mind the negative sign for sin's integral!
sec², csc²
∫ sec²x dx = tan x + C, ∫ csc²x dx = −cot x + C
Inverse of differentiation
Worked Examples
Example 1
Evaluate the indefinite integral ∫ 2x cos(x2) dx.
1
Substitute the inner function x2 as u, so du = 2x dx.
u = x2, du = 2x dx
2
Integrate in the u-world, then convert back to x.
∫ cos u du = sin u + C = sin(x2) + C
sin(x2) + C
The key to substitution is letting u be the inner function. Spotting du = g′(x)dx in the integrand is the signal.
Example 2
Evaluate ∫ x ex dx.
1
By LIATE, take u = x (algebraic) and dv = ex dx, so du = dx and v = ex.
u = x, dv = ex dx ⇒ du = dx, v = ex
2
Apply integration by parts ∫u dv = uv − ∫v du.
x ex − ∫ ex dx = x ex − ex + C
(x − 1)ex + C
In by-parts, choose u as the factor that simplifies when differentiated (LIATE order). Polynomials are usually u.
Wrap-up
Choosing Method
composite → substitution, product → IBP, radical → trig sub, rational → partial fractions
Four guidelines based on the form of the integrand
CSAT-style
What is the definite integral ∫01 2x(x2+1)3 dx?
154
4
174
152
8
154
1
Substitute u = x2+1 and change the limits too (x: 0→1 ⇒ u: 1→2).
u = x2+1, x: 0→1 ⇒ u: 1→2
2
Evaluate the definite integral in u.
12 u3 du = 24 − 144 = 154
🎯 Exam Points
①Substitution: choose u (the inner function)
②IBP: choose u by LIATE
③Trig sub: √(a²−x²)→sinθ, √(x²+a²)→tanθ, √(x²−a²)→secθ
④Partial fractions: factor denominator, solve for coefficients
⑤Always verify: differentiate your answer to recover the integrand
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