high school 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 names the strategy. Drag the upper-limit slider and the x-world turns into the u-world, and the area under xeˣ builds up.

Substitution

Upper limit2

🔄 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

Upper limit1.5

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, √(x²+a²) → x = a tan 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. Multiply by (x−a)(x−b), then substitute to find A, B

💡 Strategy

①Composite form → substitution

②Product of functions → integration by parts

③Radicals → trig substitution

④Rational functions → partial fractions

⑤After integrating, differentiate to check you get the integrand back

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.

Substitute the inner function x² as u, so du = 2x dx.

u = x², du = 2x dx

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 e^x dx.

By LIATE, take u = x (algebraic) and dv = e^x dx, so du = dx and v = e^x.

u = x, dv = e^x dx ⇒ du = dx, v = e^x

Apply integration by parts ∫u dv = uv − ∫v du.

x e^x − ∫ e^x dx = x e^x − e^x + C

▸ (x − 1)e^x + 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

exam-style

What is the definite integral ∫₀¹ 2x(x²+1)³ dx?

①154

②4

③174

④152

⑤8

▸ ① 154

Substitute u = x²+1 and change the limits too (x: 0→1 ⇒ u: 1→2).

u = x²+1, x: 0→1 ⇒ u: 1→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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