Integrals

Differentiation converts a function into its rate of change. Integration reverses this process. If F(x) is a function such that F′(x) = f(x), then F is called an antiderivative (or primitive) of f. Since the derivative of any constant is zero, if F(x) is one antiderivative of f(x), then F(x) + C is also an antiderivative of f(x) for every real number C. This entire family of functions is denoted by the indefinite integral ∫f(x) dx = F(x) + C, where C is called the constant of integration and f(x) is called the integrand.

Geometrically, y = F(x) + C represents a family of curves obtained by shifting any one member vertically. All members of this family have parallel tangents at points with the same x-coordinate, because the slope F′(x) = f(x) is the same for every value of C.

Two properties make integration manageable: it is linear. That is, the integral of a sum is the sum of the integrals, ∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx, and a constant multiplier can be pulled out, ∫k f(x) dx = k∫f(x) dx. Using these two rules together with a short list of standard integrals (obtained simply by reversing familiar derivative formulas), a large number of functions can be integrated term by term.

Worked Example: Evaluate ∫(3x² + 2sec²x − 7/√(1−x²)) dx.

Using linearity, integrate each term separately using the standard results ∫xⁿ dx = xⁿ⁺¹/(n+1) + C, ∫sec²x dx = tan x + C, and ∫dx/√(1−x²) = sin^-1x + C:

1. ∫3x² dx = 3·(x³/3) = x³
2. ∫2sec²x dx = 2tan x
3. ∫−7/√(1−x²) dx = −7sin^-1x

Adding these and writing a single constant of integration: x³ + 2tan x − 7sin^-1x + C. This is the required indefinite integral, and it can be checked instantly by differentiating the answer and confirming it returns the original integrand.

Many integrands are not directly on the standard list but become standard after a change of variable. If the integral has the form ∫f(g(x))·g′(x) dx, substitute t = g(x), so that dt = g′(x) dx. The integral becomes ∫f(t) dt, which can (hopefully) be evaluated directly; the final answer is then converted back from t to x.

Worked Example 1: Evaluate ∫2x sin(x²+1) dx.

Let t = x² + 1, so dt = 2x dx. The integral becomes ∫sin t dt = −cos t + C = −cos(x²+1) + C.

Worked Example 2 (deriving a new standard result): Evaluate ∫tan x dx.

Write tan x = sin x/cos x. Let t = cos x, so dt = −sin x dx, i.e. sin x dx = −dt. Then:

∫(sin x/cos x) dx = ∫(−dt/t) = −ln|t| + C = −ln|cos x| + C = ln|sec x| + C.

The same idea, rewriting in terms of sin x and cos x first, also gives the integral of cot x by an identical substitution t = sin x. The integrals of sec x and cosec x need one extra algebraic step — multiplying the integrand by (sec x + tan x)/(sec x + tan x), or by (cosec x + cot x)/(cosec x + cot x) — before a substitution turns them into a standard form. These four results are used constantly in later problems and should be committed to memory rather than re-derived each time.

A set of six standard forms, derived using partial fractions or trigonometric substitution, covers integrals whose denominator involves x² ± a² or its square root. To use them on a general quadratic ax² + bx + c, first complete the square so the quadratic is written as (x + p)² ± q² for some constants p and q, and then match it to the appropriate standard form.

Worked Example 1 (completing the square): Evaluate ∫dx/(x²+2x+5).

Complete the square: x² + 2x + 5 = (x+1)² + 4 = (x+1)² + 2². Using ∫dx/(x²+a²) = (1/a)tan^-1(x/a) + C with x replaced by (x+1) and a = 2, the integral equals (1/2)tan^-1((x+1)/2) + C.

When the integrand is a rational function P(x)/Q(x) that does not match these forms directly, it is decomposed into partial fractions: simpler terms with linear or irreducible quadratic denominators, each of which can then be integrated separately (if the degree of P is not less than that of Q, divide first).

Worked Example 2 (partial fractions): Evaluate ∫dx/((x+1)(x+2)).

Write 1/((x+1)(x+2)) = A/(x+1) + B/(x+2). Multiplying through: 1 = A(x+2) + B(x+1). Putting x = −1 gives A = 1; putting x = −2 gives B = −1. So the integral becomes ∫[1/(x+1) − 1/(x+2)] dx = ln|x+1| − ln|x+2| + C = ln|(x+1)/(x+2)| + C.

When the integrand is a product of two different types of functions, integration by parts is used: ∫u·v dx = u∫v dx − ∫[u′·(∫v dx)] dx. Here u is called the first function and v the second function. Choosing u correctly is essential; the preference order ILATE (Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential) generally tells us which factor to treat as u — whichever type appears earliest in this order should be u, since it usually simplifies on differentiation.

Worked Example 1: Evaluate ∫x e^x dx. By ILATE, the algebraic factor x is chosen as u (it comes before the exponential), and e^x dx is v dx. So u = x, dv = e^xdx, giving v = e^x. Then ∫x e^x dx = x e^x − ∫e^x dx = x e^x − e^x + C = e^x(x−1) + C.

A special and very useful shortcut arises when the integrand has the form e^x[f(x) + f′(x)]: applying the product rule in reverse shows that ∫e^x[f(x)+f′(x)] dx = e^xf(x) + C, with no integration by parts needed at all once this pattern is spotted.

Worked Example 2: Evaluate ∫e^x(1/x − 1/x²) dx. Here f(x) = 1/x, so f′(x) = −1/x², matching the integrand exactly. So the integral is simply e^x·(1/x) + C = e^x/x + C.

The same by-parts technique, applied carefully with a trigonometric substitution, also gives three memorable results for integrating square roots of quadratics, which appear repeatedly in board and competitive exams.

Consider the region bounded by y = f(x), the x-axis, and the vertical lines x = a and x = b. Its area can be approximated by dividing [a, b] into n very thin strips of width h = (b−a)/n and adding up the areas of rectangles built on each strip. As n → ∞ (equivalently h → 0), this sum approaches the exact area, which we call the definite integral of f from a to b, written ∫_a^b f(x) dx.

Calculating this limit directly for every problem would be extremely slow. The Fundamental Theorem of Calculus avoids this entirely: if F is any antiderivative of f (that is, F′(x) = f(x)), then ∫_a^b f(x) dx = F(b) − F(a). Notice that the arbitrary constant C cancels out in this subtraction, so it is simply dropped when evaluating a definite integral.

Worked Example 1: Evaluate ∫₀² (x²+1) dx.

An antiderivative is F(x) = x³/3 + x. By the Fundamental Theorem, the integral equals F(2) − F(0) = (8/3 + 2) − (0 + 0) = 8/3 + 6/3 = 14/3.

Worked Example 2 (substitution with limit change): Evaluate ∫₀¹ x/√(1+x²) dx.

Let t = 1 + x², so dt = 2x dx. When x = 0, t = 1; when x = 1, t = 2 — the limits are changed to the new variable, so there is no need to substitute back to x at the end. The integral becomes (1/2)∫₁² dt/√t = [√t]₁² = √2 − 1.

Definite integrals obey several useful symmetry properties. The variable of integration is a dummy variable and can be renamed freely (P0). Reversing the limits of integration flips the sign of the integral (P1), and an integral with equal upper and lower limits is zero (P2). Any interior point c can be used to split the interval of integration into two pieces whose integrals add up to the original (P3).

Two properties involve reflecting x within the interval: replacing x by (a+b−x) leaves ∫_a^b f(x) dx unchanged (P4), and in the special case a = 0, replacing x by (a−x) leaves ∫₀^a f(x) dx unchanged (P5). These are extremely powerful because they can convert a difficult integrand into a different, sometimes much simpler, one over the very same interval.

Two further properties concern symmetric intervals: an integral over [0, 2a] can be related to [0, a] via f(2a−x) — it equals twice the [0, a] integral when f(2a−x) = f(x), and equals zero when f(2a−x) = −f(x) (P6). An integral over [−a, a] equals twice the integral over [0, a] if f is even, and equals 0 if f is odd (P7).

Worked Example (using P5): Evaluate I = ∫₀^π/2 √(sin x)/(√(sin x)+√(cos x)) dx.

Replace x by (π/2 − x) using property P5: I = ∫₀^π/2 √(cos x)/(√(cos x)+√(sin x)) dx. Adding this to the original expression for I:

2I = ∫₀^π/2 [√(sin x)+√(cos x)]/[√(sin x)+√(cos x)] dx = ∫₀^π/2 1 dx = π/2.

So I = π/4, found without ever computing an antiderivative of the original integrand.

Quick odd-function check: ∫_-2² x³cos x dx = 0 immediately, since x³ is odd and cos x is even, making the product an odd function integrated over a symmetric interval [−2, 2] (property P7).