Unit 4: Integration I

Integration is the reverse operation to differentiation: given a rate of change, it recovers the quantity itself. This unit develops the antiderivative (indefinite integral) and the systematic techniques that turn awkward integrands into standard forms. Every method here is an act of pattern-matching — you manipulate the integrand until it matches something you already know how to reverse-differentiate.

• Notation: ∫ f(x) dx = F(x) + C, where F′(x) = f(x). Here f(x) is the integrand, dx marks the variable, F(x) is the antiderivative and C is the arbitrary constant of integration.
• Family of solutions: because d/dx (C) = 0, every antiderivative differs only by a constant, so an indefinite integral names a whole family of curves.
• Linearity: ∫ [a f(x) + b g(x)] dx = a∫ f(x) dx + b∫ g(x) dx for constants a, b.
• Standard results relied on throughout: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1), ∫ (1/x) dx = ln|x| + C, ∫ eˣ dx = eˣ + C, ∫ sin x dx = −cos x + C, ∫ cos x dx = sin x + C, ∫ sec²x dx = tan x + C.

II. Integration as an Inverse Process of Differentiation

Recovering the function from its derivative.

The whole subject rests on undoing the differentiation table entry by entry.

A. The antiderivative relationship

The integral is defined as the operation whose output differentiates back to the integrand.

• Definition: ∫ f(x) dx = F(x) + C precisely when dF/dx = f(x).
• Reading the derivative table backwards: since d/dx(sin x) = cos x, it follows at once that ∫ cos x dx = sin x + C.
• Verification is always available: any integration result can be checked by differentiating it. If ∫ f dx = F + C, then F′ must reproduce f.
• Role of C: the constant records the information lost by differentiation; a definite condition (a point on the curve) is needed to fix it.

B. Geometrical and physical meaning

The antiderivative has a concrete interpretation that motivates the algebra.

• Family of parallel curves: each value of C shifts F(x) vertically, giving curves with identical slope f(x) at every x.
• From rate to quantity: if velocity is v(t) = ds/dt, then s(t) = ∫ v(t) dt recovers displacement, with C set by the initial position.

III. Integration by Substitution

Reversing the chain rule by changing the variable.

Substitution converts an integral into a simpler one by replacing an inner expression with a new variable.

A. Principle and procedure

The method works because it undoes the chain rule d/dx F(g(x)) = F′(g(x))·g′(x).

• Core rule: put t = g(x), so dt = g′(x) dx, giving
  
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  ∫ f(g(x)) g′(x) dx = ∫ f(t) dt

• What to choose: pick t as the part whose derivative also appears (up to a constant) elsewhere in the integrand.
• Return to x: after integrating in t, substitute g(x) back so the answer is expressed in the original variable.
• Worked example:
  
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  ∫ 2x·cos(x²) dx      let t = x², dt = 2x dx
   = ∫ cos t dt = sin t + C = sin(x²) + C

B. Standard forms produced by substitution

Several textbook results are just substitution applied once.

• ∫ tan x dx = ln|sec x| + C: put t = cos x, dt = −sin x dx, giving ∫ −dt/t = −ln|t|.
• ∫ f′(x)/f(x) dx = ln|f(x)| + C: any integrand that is a derivative over its own function integrates to a logarithm.
• ∫ [f(x)]ⁿ f′(x) dx = [f(x)]ⁿ⁺¹/(n+1) + C for n ≠ −1.

IV. Integration Using Trigonometric Identities

Rewriting products and powers of trig functions into integrable sums.

When no direct formula fits, a trigonometric identity flattens the integrand into terms with known integrals.

A. Powers of sine and cosine

Even powers use double-angle identities to drop the exponent.

• sin²x and cos²x: apply sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2.
• Worked example:
  
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  ∫ sin²x dx = ∫ (1 − cos 2x)/2 dx
   = x/2 − (sin 2x)/4 + C

• Odd powers: split off one factor and convert the rest with sin²x + cos²x = 1, then substitute.

B. Products of sines and cosines

Product-to-sum identities remove the multiplication.

• Identities used: 2 sin A cos B = sin(A+B) + sin(A−B); 2 cos A cos B = cos(A+B) + cos(A−B); 2 sin A sin B = cos(A−B) − cos(A+B).
• Example pattern: ∫ sin 3x cos x dx = ½∫ [sin 4x + sin 2x] dx.

C. Powers of tangent and secant

Identities involving sec²x set up a clean substitution.

• Key identity: 1 + tan²x = sec²x lets ∫ tan²x dx = ∫ (sec²x − 1) dx = tan x − x + C.
• Reduction hook: because d/dx(tan x) = sec²x, factoring out sec²x invites t = tan x.

V. Integrals of Some Particular Functions

A memorised table of standard quadratic-denominator and root forms.

These results, derived once by substitution or completing the square, are applied directly as formulae.

A. The standard reciprocal-quadratic forms

Each has a fixed answer depending on the sign pattern; a is a positive constant.

∫ 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

• Distinguishing them: a difference of squares gives a logarithm; a sum of squares gives an inverse tangent.

B. The standard surd forms

Square roots in the denominator produce inverse-trig or logarithmic answers.

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

• Symbol note: a > 0, and each formula assumes the radicand stays positive over the interval.

C. Completing the square

Any quadratic ax² + bx + c is forced into one of the standard shapes before the table is used.

• Method: rewrite x² + bx + c = (x + b/2)² + (c − b²/4), then substitute t = x + b/2.
• Example:
  
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  ∫ dx/(x² + 4x + 8) = ∫ dt/(t² + 4)   with t = x + 2
   = ½ tan⁻¹((x+2)/2) + C

• Linear numerators: split ∫ (px + q)/(quadratic) dx into a piece proportional to the derivative of the denominator (a log) plus a constant piece (a standard form).

VI. Integration by Partial Fractions

Splitting a rational function into simple fractions that integrate term by term.

A ratio of polynomials P(x)/Q(x) is decomposed into a sum of fractions each matching a standard integral.

A. Setting up the decomposition

The form of the split depends on how the denominator factorises.

• Proper fraction first: if deg P ≥ deg Q, divide so that a polynomial plus a proper remainder fraction results.
• Distinct linear factors: for (x−a)(x−b), write A/(x−a) + B/(x−b).
• Repeated linear factor: for (x−a)², write A/(x−a) + B/(x−a)².
• Irreducible quadratic factor: for (x² + bx + c), write (Ax + B)/(x² + bx + c).

B. Finding the constants and integrating

Once split, each piece is a form already covered above.

• Solving for coefficients: multiply through by Q(x) and either equate coefficients of like powers or substitute convenient x-values (e.g. the roots) to isolate each constant.
• Integrating the pieces: ∫ A/(x−a) dx = A ln|x−a|; a 1/(x−a)² term gives −A/(x−a); a quadratic term yields a log plus an inverse tangent.
• Worked example:
  
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  ∫ dx/[(x−1)(x+2)]
   = ∫ [ (1/3)/(x−1) − (1/3)/(x+2) ] dx
   = (1/3) ln|x−1| − (1/3) ln|x+2| + C
   = (1/3) ln|(x−1)/(x+2)| + C

• Constants shown: solving 1 = A(x+2) + B(x−1) at x = 1 gives A = 1/3; at x = −2 gives B = −1/3.

C. When partial fractions is the right tool

The method targets integrands no substitution simplifies directly.

• Trigger: the integrand is a proper rational function whose denominator factorises.
• Payoff: it reduces an intractable ratio to a sum of logarithms and inverse tangents, each drawn from the standard tables of Sections V and II.
• Limitation: it applies only to rational functions; an irreducible denominator of high degree still requires completing the square on each quadratic block.