Calculus · Chapter 7 — Techniques of Integration

SECTION 7.1

Integration by Parts

Every differentiation rule has a matching integration rule. Integration by parts is the partner of the Product Rule, and it is the go-to method whenever an integrand is a product of two functions of different types.

Integration by parts

A method for integrating a product, derived by reversing the Product Rule. Choose one factor to differentiate (call it \(u\)) and one to integrate (call it \(dv\)); the integral is then rewritten in terms of a hopefully simpler integral.

Integration by parts

\[ \int u\,dv = uv - \int v\,du \]

For definite integrals, evaluate the boundary term and the remaining integral over the same limits.

Definite form

\[ \int_a^b u\,dv = \big[\,uv\,\big]_a^b - \int_a^b v\,du \]

Choosing \(u\) and \(dv\)

Pick \(u\) to be the factor that becomes simpler (or no worse) when differentiated, and \(dv\) to be a factor you can readily integrate. A wrong choice usually produces a harder integral than the one you started with — a useful signal to swap.

EXAMPLE 1 A polynomial times a trig function

Find \( \displaystyle\int x\sin x\,dx \).

Solution

Let \( u = x \), \( dv = \sin x\,dx \). Then \( du = dx \) and \( v = -\cos x \).

Apply the formula: \( \int x\sin x\,dx = -x\cos x + \int \cos x\,dx \).

\( = -x\cos x + \sin x + C \). (Differentiate to confirm it returns \(x\sin x\).)

EXAMPLE 2 Integrating the logarithm

Evaluate \( \displaystyle\int \ln x\,dx \).

Solution

With little choice, let \( u = \ln x \), \( dv = dx \), so \( du = \tfrac{1}{x}\,dx \), \( v = x \).

\( \int \ln x\,dx = x\ln x - \int x\cdot\tfrac{1}{x}\,dx = x\ln x - \int dx \).

\( = x\ln x - x + C \).

EXAMPLE 3 Applying parts twice

Find \( \displaystyle\int t^2 e^t\,dt \).

Solution

Let \( u = t^2 \), \( dv = e^t\,dt \): \( \int t^2 e^t\,dt = t^2 e^t - 2\int t e^t\,dt \).

The new integral is simpler but not immediate, so apply parts again with \( u = t \): \( \int t e^t\,dt = t e^t - e^t + C \).

\( \int t^2 e^t\,dt = t^2 e^t - 2t e^t + 2e^t + C \).

EXAMPLE 4 A self-referencing integral

Evaluate \( \displaystyle\int e^x\sin x\,dx \).

Solution

Parts twice returns the original integral on the right side, giving \( \int e^x\sin x\,dx = -e^x\cos x + e^x\sin x - \int e^x\sin x\,dx \).

Treat it as an equation: add \( \int e^x\sin x\,dx \) to both sides and divide by 2.

\( \int e^x\sin x\,dx = \tfrac{1}{2}e^x\big(\sin x - \cos x\big) + C \).

Reduction formulas

Repeated integration by parts can express a power integral in terms of a lower power, yielding a reduction formula that is applied recursively.

Reduction formula for \(\sin^n x\) (\(n \ge 2\))

\[ \int \sin^n x\,dx = -\frac{1}{n}\cos x\,\sin^{n-1}x + \frac{n-1}{n}\int \sin^{n-2}x\,dx \]

Instructor Note

Memorise the picture, not the letters: "differentiate the algebra, integrate the wave." For \(x^n\) times \(\sin\), \(\cos\), or \(e^x\), \(u\) is almost always the power of \(x\). For a lone \(\ln x\) or \(\arctan x\), let that be \(u\) and \(dv = dx\).

SECTION 7.2

Trigonometric Integrals

Here trigonometric identities convert awkward products of sines, cosines, secants, and tangents into forms that yield to a simple substitution.

Odd-power strategy

When sine or cosine appears to an odd power, peel off one factor and convert the remaining even power using \( \sin^2 x + \cos^2 x = 1 \). The leftover single factor becomes the differential for a substitution.

Pythagorean identity

\[ \sin^2 x + \cos^2 x = 1 \]

EXAMPLE 1 Odd power of cosine

Evaluate \( \displaystyle\int \cos^3 x\,dx \).

Solution

Write \( \cos^3 x = (1-\sin^2 x)\cos x \) and let \( u = \sin x \), \( du = \cos x\,dx \).

\( \int (1-u^2)\,du = u - \tfrac{1}{3}u^3 + C \).

\( = \sin x - \tfrac{1}{3}\sin^3 x + C \).

Even-power strategy (half-angle)

When both powers are even, the peeling trick fails. Reduce the degree instead with the half-angle identities, which trade a squared term for a first-power cosine of double the angle.

Half-angle identities

\[ \sin^2 x = \tfrac{1}{2}\big(1-\cos 2x\big), \qquad \cos^2 x = \tfrac{1}{2}\big(1+\cos 2x\big) \]

EXAMPLE 3 Even power via half-angle

Evaluate \( \displaystyle\int_0^{\pi} \sin^2 x\,dx \).

Solution

Replace \( \sin^2 x \) with \( \tfrac{1}{2}(1-\cos 2x) \).

\( \tfrac{1}{2}\int_0^{\pi}(1-\cos 2x)\,dx = \tfrac{1}{2}\big[x - \tfrac{1}{2}\sin 2x\big]_0^{\pi} \).

\( = \tfrac{\pi}{2} \).

Secant–tangent strategy

For \( \int \tan^m x \sec^n x\,dx \): if the secant power is even, save \( \sec^2 x \) for the differential and convert the rest with \( \sec^2 x = 1 + \tan^2 x \). If the tangent power is odd, save \( \sec x\tan x \) and convert the rest with \( \tan^2 x = \sec^2 x - 1 \).

Integral of secant

\[ \int \sec x\,dx = \ln\big|\sec x + \tan x\big| + C \]

EXAMPLE 8 An odd power of secant by parts

Find \( \displaystyle\int \sec^3 x\,dx \).

Solution

Integrate by parts with \( u = \sec x \), \( dv = \sec^2 x\,dx \); the integral reappears and is solved algebraically.

\( \int \sec^3 x\,dx = \tfrac{1}{2}\big(\sec x\tan x + \ln|\sec x + \tan x|\big) + C \).

Product-to-sum for mixed frequencies

Integrals like \( \int \sin mx\cos nx\,dx \) are easiest after converting the product to a sum of single trig terms, each integrated directly.

Product identities

\[ \sin A\cos B = \tfrac{1}{2}\big[\sin(A-B) + \sin(A+B)\big] \]

SECTION 7.3

Trigonometric Substitution

When an integrand contains a square root of a quadratic, replacing the variable with a trig function lets the Pythagorean identity dissolve the root.

Inverse substitution

Rather than setting a new variable equal to part of the integrand, here the old variable is written as a trig function of a new one. Restricting the angle to a standard interval keeps the substitution one-to-one and the root non-negative.

The three standard substitutions

\[ \sqrt{a^2 - x^2}:\ x = a\sin\theta \qquad \sqrt{a^2 + x^2}:\ x = a\tan\theta \qquad \sqrt{x^2 - a^2}:\ x = a\sec\theta \]

FIGURE 7.3-1 Reference triangle for \( x = 3\sin\theta \): the angle's sine is \( x/3 \), so the adjacent side is \( \sqrt{9-x^2} \). Redrawn for MechForge.

EXAMPLE 1 Sine substitution

Evaluate \( \displaystyle\int \frac{\sqrt{9 - x^2}}{x^2}\,dx \).

Solution

Let \( x = 3\sin\theta \), \( dx = 3\cos\theta\,d\theta \); then \( \sqrt{9-x^2} = 3\cos\theta \).

The integral becomes \( \int \cot^2\theta\,d\theta = \int(\csc^2\theta - 1)\,d\theta = -\cot\theta - \theta \).

Read \( \cot\theta = \tfrac{\sqrt{9-x^2}}{x} \) from the reference triangle and convert back.

\( = -\dfrac{\sqrt{9-x^2}}{x} - \sin^{-1}\!\Big(\dfrac{x}{3}\Big) + C \).

EXAMPLE 2 Area of an ellipse

Find the area enclosed by \( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \).

Solution

By symmetry the area is \( 4\cdot\dfrac{b}{a}\displaystyle\int_0^a \sqrt{a^2 - x^2}\,dx \); substitute \( x = a\sin\theta \).

The definite integral changes limits to \( 0 \to \tfrac{\pi}{2} \) and evaluates with a half-angle identity.

Area \( = \pi a b \). Setting \( a = b = r \) recovers \( \pi r^2 \) for a circle.

Completing the square first

A general quadratic under a root (e.g. \( \sqrt{3 + 2x - x^2} \)) is reduced to one of the three standard forms by completing the square, then a shift \( u = x - h \) followed by the matching trig substitution.

Instructor Note

Always check for a plain \(u\)-substitution first. If the integrand has an \(x\) sitting outside the root (like \( \int \tfrac{x}{\sqrt{x^2+4}}\,dx \)), the easy substitution \( u = x^2+4 \) beats trig substitution every time. Reach for sine/tan/sec only when no lone matching factor is available.

SECTION 7.4

Rational Functions & Partial Fractions

Any rational function can be integrated by splitting it into a sum of simple fractions whose integrals are already known.

Proper vs. improper

A rational function is proper when the numerator's degree is less than the denominator's. If it is improper, first perform polynomial long division so the leftover fraction is proper before decomposing.

The four cases of decomposition

Case I — distinct linear factors: one term \( \tfrac{A}{ax+b} \) each. Case II — a repeated linear factor \( (ax+b)^r \): terms with denominators \( (ax+b), (ax+b)^2, \dots, (ax+b)^r \). Case III — an irreducible quadratic factor: a term \( \tfrac{Ax+B}{ax^2+bx+c} \). Case IV — a repeated irreducible quadratic: one such term per power.

Useful standard integral

\[ \int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C \]

EXAMPLE 2 Distinct linear factors

Evaluate \( \displaystyle\int \frac{x^2 + 2x - 1}{2x^3 + 3x^2 - 2x}\,dx \).

Solution

Factor the denominator: \( x(2x-1)(x+2) \). Write as \( \tfrac{A}{x} + \tfrac{B}{2x-1} + \tfrac{C}{x+2} \).

Clearing denominators and matching coefficients gives \( A = \tfrac{1}{2} \), \( B = \tfrac{1}{5} \), \( C = -\tfrac{1}{10} \).

\( = \tfrac{1}{2}\ln|x| + \tfrac{1}{10}\ln|2x-1| - \tfrac{1}{10}\ln|x+2| + K \).

EXAMPLE 5 An irreducible quadratic factor

Evaluate \( \displaystyle\int \frac{2x^2 - x + 4}{x^3 + 4x}\,dx \).

Solution

Factor \( x(x^2+4) \) and decompose as \( \tfrac{1}{x} + \tfrac{x-1}{x^2+4} \).

Split the second term and integrate each piece (log, log, and arctangent).

\( = \ln|x| + \tfrac{1}{2}\ln(x^2+4) - \tfrac{1}{2}\tan^{-1}\!\big(\tfrac{x}{2}\big) + K \).

Rationalizing substitution

An integrand containing \( \sqrt[n]{g(x)} \) can sometimes be turned into a rational function by the substitution \( u = \sqrt[n]{g(x)} \), after which partial fractions applies.

Instructor Note

Use the "plug-in" shortcut for distinct linear factors: after clearing denominators, substitute the root of each factor to isolate one constant at a time. It's far faster than solving the full coefficient system, and it's exam-friendly.

SECTION 7.5

Strategy for Integration

Unlike differentiation, integration has no algorithm that always works. This section is a strategy for recognising which technique a given integral calls for.

The four-step strategy

1. Simplify the integrand with algebra or trig identities. 2. Look for an obvious substitution — a function whose differential also appears. 3. Classify by form: trig products → §7.2 methods; rational → partial fractions; a power times a transcendental → parts; a quadratic root → trig substitution. 4. Try again — fundamentally there are only two engines, substitution and parts, sometimes used in combination.

FIGURE 7.5-1 The integration strategy as a decision flow. Original diagram for MechForge.

Not every integral is elementary

Some continuous functions have no antiderivative expressible through elementary functions. Integrals such as \( \int e^{x^2}\,dx \), \( \int \tfrac{\sin x}{x}\,dx \), and \( \int \tfrac{1}{\ln x}\,dx \) exist but cannot be written in closed form — a limitation of notation, not of the integral itself.

SECTION 7.6

Tables & Computer Algebra Systems

Reference tables and computer algebra systems (CAS) extend reach, but both require matching the integral to a stored form — often after a substitution.

Using a table of integrals

Integrals rarely appear in exactly the listed form. The skill is transforming the given integral — by substitution or algebra — into one of the catalogued patterns, then reading off the result.

Reading CAS output critically

A CAS returns a particular antiderivative: it omits the constant of integration and the absolute-value signs that a human would supply. Different systems can also return algebraically equivalent answers that look entirely different. Verify, add \(+C\), and restore domains where needed.

EXAMPLE 2 Matching a table form via substitution

Use a table to find \( \displaystyle\int \frac{x^2}{\sqrt{5 - 4x^2}}\,dx \).

Solution

Substitute \( u = 2x \) so the radical becomes \( \sqrt{5 - u^2} \), matching a standard \( \sqrt{a^2 - u^2} \) entry with \( a = \sqrt{5} \).

\( = -\dfrac{x}{8}\sqrt{5 - 4x^2} + \dfrac{5}{16}\sin^{-1}\!\Big(\dfrac{2x}{\sqrt 5}\Big) + C \).

Instructor Note

A CAS is a calculator, not a tutor. If two answers disagree, they're usually equal up to a constant — check by differentiating both. Knowing the hand methods is what lets you tell a real discrepancy from a cosmetic one.

SECTION 7.7

Approximate Integration

When an antiderivative cannot be found, or the function is known only from data, numerical rules estimate the definite integral to controllable accuracy.

Why approximate

Two situations force estimation: integrands with no elementary antiderivative (e.g. \( e^{x^2} \)), and functions supplied only as instrument readings or a table of values with no formula at all.

Midpoint Rule

\[ \int_a^b f(x)\,dx \approx \Delta x\,\big[\,f(\bar x_1) + f(\bar x_2) + \cdots + f(\bar x_n)\,\big],\quad \Delta x = \frac{b-a}{n} \]

Trapezoidal Rule

\[ \int_a^b f(x)\,dx \approx \frac{\Delta x}{2}\big[\,f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\,\big] \]

Simpson's Rule (\(n\) even)

\[ \int_a^b f(x)\,dx \approx \frac{\Delta x}{3}\big[\,f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n-1}) + f(x_n)\,\big] \]

FIGURE 7.7-1 The Trapezoidal Rule replaces the curve with straight chords across each subinterval. Original illustration for MechForge.

Comparing accuracy

Trapezoidal and Midpoint errors shrink by roughly a factor of 4 when \(n\) doubles, and they have opposite signs — the Midpoint error being about half the Trapezoidal error. Simpson's Rule, which fits parabolas, is far more accurate: its error falls by a factor of about 16 when \(n\) doubles. Simpson's estimate is a weighted blend, \( S_{2n} = \tfrac{1}{3}T_n + \tfrac{2}{3}M_n \).

Error bound — Simpson's Rule

If \( \big|f^{(4)}(x)\big| \le K \) for all \( a \le x \le b \), then the error \( E_S \) in Simpson's Rule satisfies

\[ |E_S| \le \frac{K(b-a)^5}{180\,n^4} \]

EXAMPLE 4 Simpson's Rule in action

Approximate \( \displaystyle\int_1^2 \frac{1}{x}\,dx \) using Simpson's Rule with \( n = 10 \).

Solution

With \( \Delta x = 0.1 \), apply the \( 1,4,2,4,\dots,4,1 \) coefficient pattern to \( f(x) = 1/x \).

\( S_{10} \approx 0.693150 \), against the true value \( \ln 2 \approx 0.693147 \) — accurate to five decimals.

Instructor Note

Remember Simpson's coefficient rhythm: endpoints get 1, then alternate 4, 2, 4, 2, …, ending in 4, 1. It only works for an even number of subintervals. When you're handed a table of measured data, this is the rule to use.

SECTION 7.8

Improper Integrals

The definite integral is extended to infinite intervals and to integrands with infinite discontinuities, each defined as a limit of ordinary integrals.

Type 1 — infinite interval

An integral over an unbounded interval is defined as the limit of integrals over finite pieces. If the limit is a finite number the integral converges; otherwise it diverges.

Type 1 definition

\[ \int_a^{\infty} f(x)\,dx = \lim_{t\to\infty}\int_a^{t} f(x)\,dx \]

Type 2 — discontinuous integrand

If the integrand has a vertical asymptote at an endpoint, the integral is defined by taking a limit as the bound approaches that point. A discontinuity inside the interval is handled by splitting at the trouble point.

EXAMPLE 1 A divergent case

Determine whether \( \displaystyle\int_1^{\infty} \frac{1}{x}\,dx \) converges.

Solution

\( \displaystyle\lim_{t\to\infty}\int_1^{t}\frac{1}{x}\,dx = \lim_{t\to\infty}\ln t \), which grows without bound.

The integral diverges.

The \(p\)-integral test

\[ \int_1^{\infty} \frac{1}{x^p}\,dx \ \text{converges if } p > 1, \ \text{diverges if } p \le 1 \]

EXAMPLE 3 Convergence over the whole line

Evaluate \( \displaystyle\int_{-\infty}^{\infty} \frac{1}{1 + x^2}\,dx \).

Solution

Split at \( 0 \); each half is an arctangent limit, contributing \( \tfrac{\pi}{2} \).

\( = \pi \) — interpretable as the total area under the curve.

Comparison Theorem

Suppose \( f \) and \( g \) are continuous with \( f(x) \ge g(x) \ge 0 \) for \( x \ge a \).

(a) If \( \displaystyle\int_a^{\infty} f(x)\,dx \) converges, then \( \displaystyle\int_a^{\infty} g(x)\,dx \) converges.

(b) If \( \displaystyle\int_a^{\infty} g(x)\,dx \) diverges, then \( \displaystyle\int_a^{\infty} f(x)\,dx \) diverges.

EXAMPLE 9 Convergence without evaluation

Show that \( \displaystyle\int_0^{\infty} e^{-x^2}\,dx \) converges.

Solution

The antiderivative is not elementary, so compare instead. For \( x \ge 1 \), \( e^{-x^2} \le e^{-x} \).

Since \( \int_1^{\infty} e^{-x}\,dx \) converges, the tail converges by comparison; the bounded part is an ordinary integral.

Therefore \( \int_0^{\infty} e^{-x^2}\,dx \) converges (its value is \( \tfrac{\sqrt{\pi}}{2} \)).

Instructor Note

Never integrate across a hidden asymptote as if it were ordinary — a sign error there can produce a finite "answer" for a divergent integral. Scan the integrand for points where the denominator vanishes before you compute. The Comparison Theorem is your friend when you only need to know whether, not what.