Integrals

🎯 Key Points

• ∫xⁿdx = xⁿ⁺¹/(n+1)+C works for ALL n except n=-1, where instead ∫1/x dx = ln|x|+C - the power rule has exactly one exception, and it's frequently tested
• Integration by parts (ILATE): pick u in order Inverse > Log > Algebraic > Trig > Exponential - whichever function type appears FIRST in this list becomes u, the rest is dv
• Definite integral ∫ₐᵇf(x)dx = F(b)-F(a) represents the NET signed area (area below the x-axis counts negative) - don't assume it always equals the visual/absolute area
• Substitution method works when the integrand contains a function and (a constant multiple of) its derivative together - spotting this pairing is the key skill, not the algebra itself

Integration

Integration is the reverse of differentiation. Used to find areas, volumes, and accumulations.

Indefinite Integrals

• ∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ -1)
• ∫1/x dx = ln|x| + C
• ∫eˣ dx = eˣ + C
• ∫aˣ dx = aˣ/ln a + C
• ∫sin x dx = -cos x + C
• ∫cos x dx = sin x + C
• ∫sec²x dx = tan x + C
• ∫cosec²x dx = -cot x + C
• ∫sec x tan x dx = sec x + C
• ∫1/√(1-x²) dx = sin⁻¹x + C
• ∫1/(1+x²) dx = tan⁻¹x + C

Definite Integrals

∫ₐᵇ f(x)dx = F(b) - F(a) where F is the antiderivative. Represents area between the curve and x-axis from a to b.

Integration Methods

• Substitution: u = g(x), replace and simplify
• Integration by parts: ∫uv dx = u∫v dx - ∫(u' ∫v dx) dx (ILATE rule for u: Inverse, Log, Algebraic, Trig, Exponential)
• Partial fractions: split rational functions into simpler fractions

Applications

• Area under curve: ∫ₐᵇ f(x) dx
• Area between curves: ∫ₐᵇ [f(x) - g(x)] dx
• Volume of revolution (x-axis): π∫ₐᵇ y² dx

Integrals of Some Special Forms

These standard results (each with +C) should be memorised, since most substitution and completing-the-square problems reduce to them:

• ∫dx/(x² - a²) = (1/2a) ln|(x - a)/(x + a)| + C
• ∫dx/(a² - x²) = (1/2a) ln|(a + x)/(a - x)| + C
• ∫dx/(x² + a²) = (1/a) tan⁻¹(x/a) + C
• ∫dx/√(a² - x²) = sin⁻¹(x/a) + C
• ∫dx/√(x² + a²) = ln|x + √(x² + a²)| + C
• ∫dx/√(x² - a²) = ln|x + √(x² - a²)| + C
• ∫√(a² - x²) dx = (x/2)√(a² - x²) + (a²/2) sin⁻¹(x/a) + C
• ∫√(x² + a²) dx = (x/2)√(x² + a²) + (a²/2) ln|x + √(x² + a²)| + C
• ∫√(x² - a²) dx = (x/2)√(x² - a²) - (a²/2) ln|x + √(x² - a²)| + C

Quadratics of the form ax² + bx + c are first reduced to one of these standard shapes by completing the square, e.g. x² + 4x + 8 = (x + 2)² + 2².

Integration by Partial Fractions

A proper rational function P(x)/Q(x) (degree of P less than degree of Q) is split into simpler fractions before integrating. Standard decompositions:

• Distinct linear factors: (px + q) / [(x - a)(x - b)] = A/(x - a) + B/(x - b)
• Repeated linear factor: (px + q) / (x - a)² = A/(x - a) + B/(x - a)²
• Irreducible quadratic factor: (px + q) / [(x - a)(x² + bx + c)] = A/(x - a) + (Bx + C)/(x² + bx + c)

If P(x)/Q(x) is improper (degree of P ≥ degree of Q), first divide to write it as a polynomial plus a proper fraction, then apply partial fractions to the proper part.

Properties of Definite Integrals

• ∫ₐᵇ f(x)dx = ∫ₐᵇ f(t)dt - the variable of integration is a dummy variable.
• ∫ₐᵇ f(x)dx = -∫ᵇᵃ f(x)dx; in particular ∫ₐᵃ f(x)dx = 0.
• ∫ₐᵇ f(x)dx = ∫ₐᶜ f(x)dx + ∫꜀ᵇ f(x)dx (splitting the interval at an interior point c).
• ∫ₐᵇ f(x)dx = ∫ₐᵇ f(a + b - x)dx (King's property).
• ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a - x)dx.
• ∫₀²ᵃ f(x)dx = 2∫₀ᵃ f(x)dx if f(2a - x) = f(x), and = 0 if f(2a - x) = -f(x).
• ∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx if f is even, and = 0 if f is odd.

Integration as the Limit of a Sum

The definite integral is defined from first principles as the limiting value of a sum of areas of thin rectangles (a Riemann sum):

∫ₐᵇ f(x)dx = lim(n→∞) h [f(a) + f(a + h) + f(a + 2h) + ... + f(a + (n-1)h)], where h = (b - a)/n.

Equivalently, ∫ₐᵇ f(x)dx = lim(h→0) h Σ f(a + rh), with r running from 0 to n - 1. This is the rigorous meaning behind "area under a curve"; the Fundamental Theorem of Calculus (∫ₐᵇ f(x)dx = F(b) - F(a)) is what lets us evaluate it without computing the sum directly.

🚀 JEE Advanced Edge

King's rule for definite integrals: ∫ₐᵇ f(x)dx = ∫ₐᵇ f(a+b-x)dx - replacing x with (a+b-x) leaves the value of a definite integral unchanged, because this substitution just reverses the direction of traversal across [a,b]. This is the standard JEE trick for integrals that look impossible directly but simplify beautifully when added to their own "King's rule" mirror image.

Even/odd function shortcuts over symmetric limits: For ∫₋ₐᵃ f(x)dx: if f is EVEN (f(-x)=f(x)), the integral equals 2∫₀ᵃf(x)dx; if f is ODD (f(-x)=-f(x)), the integral is exactly 0 - checking symmetry BEFORE attempting to integrate can solve a problem in one line instead of a lengthy computation.

Worked problem: Evaluate ∫₀^π/2 sinx/(sinx+cosx) dx using King's rule. Approach: Let I = ∫₀^π/2 sinx/(sinx+cosx)dx. By King's rule (a=0,b=π/2, so a+b-x=π/2-x): I = ∫₀^π/2 sin(π/2-x)/[sin(π/2-x)+cos(π/2-x)]dx = ∫₀^π/2 cosx/(cosx+sinx)dx. Adding both expressions for I: 2I = ∫₀^π/2 [sinx+cosx]/(sinx+cosx)dx = ∫₀^π/2 1 dx = π/2. So I = π/4.

Worked Example: Integration by Substitution

Evaluate ∫ (2x) / (x² + 1) dx.

Let u = x² + 1, so du = 2x dx. The integral becomes ∫ du/u = ln|u| + C = ln(x² + 1) + C. When the numerator is the derivative of the denominator (or a multiple of it), substitution immediately converts it to ∫du/u.

Worked Example: Integration by Parts

Evaluate ∫ x eˣ dx.

Use the ILATE order (Inverse trig, Log, Algebraic, Trig, Exponential) - choose u = x (algebraic) and dv = eˣ dx, so du = dx and v = eˣ.

∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C. ILATE tells you which factor to differentiate and which to integrate.

Integration by Substitution

• Integration is the inverse operation of differentiation: if d/dx[F(x)] = f(x) then the integral of f(x) dx = F(x) + C, where C is the constant of integration.
• Substitution replaces the variable to simplify: putting t = g(x), dt = g'(x) dx turns the integral of f(g(x))·g'(x) dx into the integral of f(t) dt.
• A very useful case: the integral of f'(x)/f(x) dx = log|f(x)| + C.
• Also the integral of tan x dx = log|sec x| + C and the integral of cot x dx = log|sin x| + C, both proved by substitution.
• After integrating in t, substitute back to the original variable x.
• For a definite integral, either change the limits to the new variable or return to x before applying the limits.

Integration by Parts

• For a product of two functions: the integral of u·v dx = u·(the integral of v dx) − the integral of [u' · (the integral of v dx)] dx.
• Choose the first function u by the ILATE order (Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential) to keep the remaining integral simpler.
• Special result: the integral of eˣ[f(x) + f'(x)] dx = eˣ·f(x) + C.
• Some integrals (for example the integral of eˣ sin x dx) return to the original after applying parts twice, and are solved by algebra.
• Standard results such as the integral of √(a² − x²) dx and the integral of √(x² + a²) dx are derived using integration by parts.

Integration Using Trigonometric Identities

• Powers of sine and cosine are reduced with half-angle identities, e.g. sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2.
• Products such as sin A cos B are rewritten as sums using the product-to-sum formulae before integrating.
• For odd powers, split off one factor and substitute (e.g. write sin³x = sin x · (1 − cos²x)).
• Identities like 1 + tan²x = sec²x help convert integrands into directly integrable forms.
• The aim is always to turn a hard trigonometric integrand into a sum of standard integrals.

The Fundamental Theorem of Calculus

• The definite integral from a to b of f(x) dx represents the (signed) area between the curve and the x-axis over [a, b].
• First fundamental theorem: if A(x) is the area function, the integral from a to x of f(t) dt, then A'(x) = f(x).
• Second fundamental theorem (evaluation): if F is any antiderivative of f, then the integral from a to b of f(x) dx = F(b) − F(a).
• This links differentiation and integration, letting a definite integral be evaluated without summing an infinite series.
• The constant of integration cancels in F(b) − F(a), so any antiderivative may be used.
• The function f must be continuous on [a, b] for the theorem to apply directly.