Integrals · Lesson 9 of 9

Chapter Summary and Practice

“Connect every integration method and choose a justified route through mixed chapter problems.”

Learning Objectives

• Connect antiderivatives, accumulated values, and the Fundamental Theorems of Calculus. • Choose a method from the structure of the complete integrand. • Combine substitution, identities, standard forms, partial fractions, and integration by parts. • Use symmetry and sign analysis to simplify definite integrals. • Check formulas, domains, constants, intermediate steps, and the reasonableness of answers.

One chapter, two questions, several connected tools

The chapter began with two questions: how can a function be recovered from its derivative, and how can an accumulated amount beneath a curve be measured? An indefinite integral answers the first with a family of functions. A definite integral answers the second with a value over an interval. The Fundamental Theorems connect them, allowing antiderivatives to compute accumulated values and accumulation functions to recover their rates.

The techniques are connected to differentiation rules. Substitution reverses the chain rule; integration by parts reverses the product rule. Identities and partial fractions prepare an expression so those inverse rules and the standard formulas become visible. The real skill is identifying structure before carrying out algebra.

Idea	Meaning and check
Antiderivative	$F' = f$ on the interval under consideration; differentiate the proposed answer.
Indefinite integral	$\int f (x) d x = F (x) + C$ represents every antiderivative on that interval.
Linearity	Split sums and constant multiples, but do not invent product or quotient rules for integrals.
Particular antiderivative	Use a supplied function value to determine the constant.
Accumulation function	$A (x) = \int_{a}^{x} f (t) d t$ has derivative $f (x)$ under continuity.
Fixed-limit integral	$\int_{a}^{b} f (x) d x = F (b) - F (a)$ under the standard theorem assumptions; no arbitrary constant remains.
Signed and geometric area	Negative contributions can cancel positive ones; integrate an absolute value for total geometric area.

Choose a method by looking for evidence

Before integrating, inspect the integrand as a whole. A short rewrite can reveal more than immediately applying the first method that comes to mind. The following guide is a decision aid: each choice must still be justified by the actual factors, their domains, and the integral that remains.

What to look for	Useful next step	Why it helps
A sum of known derivatives	Simplify and use linearity.	Each term can be integrated independently.
An inner function and its derivative	Substitute the inner function.	The chain factor becomes the differential.
Trigonometric squares or products	Apply an appropriate identity.	A sum of standard functions or a substitution emerges.
A quadratic with a linear term	Complete the square.	A shifted standard form becomes visible.
A linear numerator over a quadratic or root	Split off a multiple of the quadratic derivative.	One part becomes immediate and the remainder is standard.
A rational function	Divide if improper, factor, then decompose.	Simple fractions give logarithms, powers, or inverse tangents.
A product with a helpful derivative	Use integration by parts.	The difficult factor simplifies or a solvable integral equation appears.
An exponential times a sum resembling a function and its derivative	Check the product-derivative pattern.	The entire integrand may be one derivative.
A definite interval with reflection or parity	Test symmetry before seeking an antiderivative.	Pairing may remove an awkward factor or give the value directly.
An absolute value	Find zeros and signs, then split.	The correct expression is integrated on each subinterval.

An integrand may need two or more rows of this guide. For example, a substitution can create a rational function, partial fractions can create a quadratic standard form, and symmetry can leave a simple substitution. Do not regard a change of method as a failure; it often reflects the nested structure of the original expression.

Reconnect the standard formulas

Organise the formulas by the differentiation relationships behind them. Powers and elementary derivatives are the starting point. Quadratic forms are obtained by scaling, shifting, factoring, or trigonometric substitution. The square-root formulas then follow from integration by parts and the reciprocal-root integrals.

Family	Core relationship
Power and exceptional reciprocal	$\int x^{n} d x = x^{n + 1} / (n + 1) + C$ for $n \neq = - 1$ ; $\int d x / x = ln ∣ x ∣ + C$ .
Elementary trigonometric derivatives	$cos x \to sin x$ ; $sin x \to - cos x$ ; $sec^{2} x \to tan x$ ; $csc^{2} x \to - cot x$ ; $sec x tan x \to sec x$ ; $csc x cot x \to - csc x$ . Add $C$ .
Inverse functions	$\int d x / 1 - x^{2} = sin^{- 1} x + C = - cos^{- 1} x + C$ as families; $\int d x / (1 + x^{2}) = tan^{- 1} x + C = - cot^{- 1} x + C$ as families.
Exponentials	$\int e^{x} d x = e^{x} + C$ ; $\int a^{x} d x = a^{x} / ln a + C$ for $a > 0$ , $a \neq = 1$ .
Remaining elementary trigonometric functions	$\int tan x d x = ln ∣ sec x ∣ + C$ ; $\int cot x d x = ln ∣ sin x ∣ + C$ ; $\int sec x d x = ln ∣ sec x + tan x ∣ + C$ ; $\int csc x d x = ln ∣ csc x - cot x ∣ + C$ .
Reciprocal quadratics	$x^{2} - a^{2}$ and $a^{2} - x^{2}$ lead to logarithmic ratios; $x^{2} + a^{2}$ leads to $tan^{- 1} (x / a) / a$ .
Reciprocal square roots	$1/ x^{2} - a^{2}$ and $1/ x^{2} + a^{2}$ lead to logarithms; $1/ a^{2} - x^{2}$ leads to $sin^{- 1} (x / a)$ .
Quadratic square roots	The antiderivative contains $x x^{2} \pm a^{2} /2$ with the corresponding signed logarithmic term, or $x a^{2} - x^{2} /2 + (a^{2} /2) sin^{- 1} (x / a)$ .
By parts and the exponential pattern	$\int u d v = uv - \int v d u$ ; $\int e^{x} (f + f') d x = e^{x} f + C$ .

In this table an arrow denotes one antiderivative, not equality between the original function and its integral. All formulas retain their interval restrictions. Also, the constants in two different antiderivative descriptions need not be the same numerical constant: the families coincide after the constants are allowed to vary.

The three radical formulas togetherLaTeX

\int x^{2} - a^{2} d x \int x^{2} + a^{2} d x \int a^{2} - x^{2} d x = \frac{x}{2} x^{2} - a^{2} - \frac{a ^{2}}{2} ln ∣ x + x^{2} - a^{2} ∣ + C, = \frac{x}{2} x^{2} + a^{2} + \frac{a ^{2}}{2} ln ∣ x + x^{2} + a^{2} ∣ + C, = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} (x / a) + C .

Take

a > 0

and use an interval where the relevant expression is real and the derivative calculation is valid. Notice the negative logarithmic sign in the first formula and the positive sign in the second.

Mixed indefinite integrals

The following examples deliberately combine decisions. Read the structure first, then follow each transformation in order. Each change should have an explicit reason, and the final answer should be expressed in the original variable with one arbitrary constant.

Example — A composite root and its derivative

Problem
Evaluate $\int cos 6 x 1 + sin 6 x d x$ on an interval where $1 + sin 6 x > 0$ .

1.Choose $u = 1 + sin 6 x$ . Its derivative is $6 cos 6 x$ , so $d u = 6 cos 6 x d x$ .
2.The integral becomes $\frac{1}{6} \int u^{1/2} d u = \frac{1}{6} \cdot \frac{2}{3} u^{3/2} + C$ .
3.The answer is $(1 + sin 6 x)^{3/2} /9 + C$ . The coefficient records both the substitution factor and the power-rule factor.

Example — Rewrite powers to expose an inner derivative

Problem
Evaluate $\int (x^{4} - x)^{1/4} / x^{5} d x$ on $x > 1$ .

1.Since $x > 1$ , write $(x^{4} - x)^{1/4} = x (1 - x^{- 3})^{1/4}$ . The integrand becomes $x^{- 4} (1 - x^{- 3})^{1/4}$ .
2.Set $u = 1 - x^{- 3}$ , so $d u = 3 x^{- 4} d x$ . The integral is $\frac{1}{3} \int u^{1/4} d u$ .
3.Thus the result is $\frac{4}{15} (1 - x^{- 3})^{5/4} + C$ . The interval restriction is needed: a fourth root of $x^{4}$ is $∣ x ∣$ , so the same rewrite cannot be silently reused for negative $x$ .

Example — Division followed by partial fractions

Problem
Evaluate $\int x^{4} / [(x - 1) (x^{2} + 1)] d x$ .

1.Polynomial division gives $x^{4} / [(x - 1) (x^{2} + 1)] = x + 1 + 1/ [(x - 1) (x^{2} + 1)]$ . Verify by multiplication: $(x + 1) (x - 1) (x^{2} + 1) = x^{4} - 1$ .
2.Write the proper remainder as $A / (x - 1) + (B x + D) / (x^{2} + 1)$ . The identity $1 = A (x^{2} + 1) + (B x + D) (x - 1)$ gives $A = 1/2$ , $B = - 1/2$ , and $D = - 1/2$ .
3.The logarithmic and inverse-tangent parts integrate separately: $\int [- x / (2 (x^{2} + 1)) - 1/ (2 (x^{2} + 1))] d x = - \frac{1}{4} ln (x^{2} + 1) - \frac{1}{2} tan^{- 1} x$ .
4.Including the polynomial and linear-factor term gives $x^{2} /2 + x + \frac{1}{2} ln ∣ x - 1∣ - \frac{1}{4} ln (x^{2} + 1) - \frac{1}{2} tan^{- 1} x + C$ .

Example — Integration by parts with cancellation

Problem
Evaluate $\int [ln (ln x) + 1/ (ln x)^{2}] d x$ for $x > 1$ .

1.Integrate the first term by parts with $u = ln (ln x)$ and $d v = d x$ . This gives $x ln (ln x) - \int d x / ln x$ .
2.Do not try to find $\int d x / ln x$ in isolation. Apply integration by parts formally to it with $u = 1/ ln x$ and $d v = d x$ . The result is $x / ln x + \int d x / (ln x)^{2}$ .
3.Substitute that relation into the original combined integral. The two occurrences of $\int d x / (ln x)^{2}$ cancel.
4.The final answer is $x ln (ln x) - x / ln x + C$ . Differentiating the complete expression checks the cancellation even though the separated integrals are not elementary.

Example — Successive substitutions after an algebraic rearrangement

Problem
Evaluate $\int [cot x + tan x] d x$ on an interval where $tan x > 0$ .

1.Let $t = tan x > 0$ , so $tan x = t^{2}$ and $d x = 2 t d t / (1 + t^{4})$ . The two root terms are $1/ t$ and $t$ .
2.The integral becomes $2 \int (1 + t^{2}) / (1 + t^{4}) d t$ . Divide numerator and denominator by $t^{2}$ to write it as $2 \int (1 + t^{- 2}) / [(t - t^{- 1})^{2} + 2] d t$ .
3.Put $y = t - t^{- 1}$ , so $d y = (1 + t^{- 2}) d t$ . The integral becomes $2 \int d y / (y^{2} + 2) = 2 tan^{- 1} (y / 2) + C$ .
4.Substituting back gives $2 tan^{- 1} [(tan x - 1) / 2 tan x] + C$ . Each substitution transformed both the expression and the differential.

Example — A trigonometric expression inside a root

Problem
Evaluate $\int sin 2 x cos 2 x / 9 - cos^{4} (2 x) d x$ .

1.Set $u = cos^{2} (2 x)$ . Then $d u = - 4 sin 2 x cos 2 x d x$ .
2.The integral is $- \frac{1}{4} \int d u / 9 - u^{2}$ .
3.Use the reciprocal-root formula to get $- \frac{1}{4} sin^{- 1} (u /3) + C = - \frac{1}{4} sin^{- 1} (cos^{2} (2 x) /3) + C$ .

Mixed definite integrals

For a definite integral, the interval is part of the problem from the beginning. Check the integrand, look for sign changes, and test reflection or parity before committing to a lengthy antiderivative. When several methods are needed, write the new limits or the interval splits explicitly.

Example — Absolute value followed by integration by parts

Problem
Evaluate $\int_{- 1}^{3/2} ∣ x sin (π x) ∣ d x$ .

1.On $[- 1, 1]$ , $x$ and $sin (π x)$ have the same sign, so their product is nonnegative. On $[1, 3/2]$ , the product is nonpositive.
2.The integral is $\int_{- 1}^{1} x sin (π x) d x - \int_{1}^{3/2} x sin (π x) d x$ .
3.Integration by parts gives $F (x) = - x cos (π x) / π + sin (π x) / π^{2}$ .
4.The first contribution is $F (1) - F (- 1) = 2/ π$ . The second, after its minus sign, is $F (1) - F (3/2) = 1/ π + 1/ π^{2}$ .
5.Therefore the total is $3/ π + 1/ π^{2}$ . The absolute value makes both contributions nonnegative.

Example — Reflection, a split, and two substitutions

Problem
For $a, b > 0$ , evaluate $I = \int_{0}^{π} x / [a^{2} cos^{2} x + b^{2} sin^{2} x] d x$ .

1.The denominator is positive and unchanged by $x \mapsto π - x$ . Adding the original and reflected forms gives $2 I = π \int_{0}^{π} d x / [a^{2} cos^{2} x + b^{2} sin^{2} x]$ .
2.The reciprocal denominator is symmetric about $π /2$ , so $I = π \int_{0}^{π /2} d x / [a^{2} cos^{2} x + b^{2} sin^{2} x]$ .
3.Split the remaining interval at $π /4$ . On $[0, π /4]$ , divide by $cos^{2} x$ and put $t = tan x$ , giving $\int_{0}^{1} d t / (a^{2} + b^{2} t^{2})$ .
4.On $[π /4, π /2]$ , divide by $sin^{2} x$ and put $u = cot x$ . Since $d u = - csc^{2} x d x$ , this gives $\int_{0}^{1} d u / (a^{2} u^{2} + b^{2})$ .
5.The two values are $tan^{- 1} (b / a) / (ab)$ and $tan^{- 1} (a / b) / (ab)$ . For positive $a, b$ , their angles add to $π /2$ .
6.Multiplying by the earlier factor $π$ gives $I = π^{2} / (2 ab)$ . The split keeps both substitutions on finite, well-defined endpoint intervals.

A useful symmetry consequenceLaTeX

g (a + b - x) = g (x) ⟹ \int_{a}^{b} xg (x) d x = \frac{a + b}{2} \int_{a}^{b} g (x) d x

Add the original integral to its reflected version. The two weights

x

and

a + b - x

add to the constant

a + b

. Assume the integrals exist.

Checking a solution from beginning to end

A calculation should pass three kinds of check. First, the transformations must preserve the original expression and its domain. Second, the antiderivative should differentiate to the integrand. Third, a definite value should agree with the sign, interval, and symmetry of the problem. A positive continuous integrand on an interval of positive length cannot have a negative integral.

For indefinite integrals, verify the chain factors and include one final $C$ . For definite integrals, check the upper-minus-lower order and omit the arbitrary constant. In a substitution verify the transformed endpoints; in partial fractions multiply the pieces back together; in integration by parts keep the minus sign and inspect the remaining integral. These checks address different possible mistakes, so one does not replace all the others.

Common mistake

Do not use a formula merely because its symbols look similar. Reciprocal roots and roots have different antiderivatives. A symmetric interval alone does not make an integral zero: the integrand must be odd and the integral must exist. A radical or logarithm rewrite also needs its domain; for example, $x^{2} = ∣ x ∣$ and not always $x$ .

Quick check

Which first step is most useful for $\int (2 x + 3) / (x^{2} + 3 x + 5) d x$ ?

Quick check

Which answer type is appropriate for $\int d x / 4 - x^{2}$ ?

Quick check

For an improper rational integrand, which step should precede partial fractions?

Quick check

What does $\int_{- 2}^{2} x^{3} cos^{2} x d x$ equal?

Quick check

Which statement correctly distinguishes $\int_{0}^{1} 2 x d x$ from $\int 2 x d x$ ?

Quick check

Why can $\int_{- 1}^{1} 1/ x d x$ not be justified as zero using the continuous odd-function rule?

Quick check

Which is a valid identity on the real line?

Practice Problems

Find $\int (3 x^{2} - 4/ x + 2 e^{x}) d x$ on $x > 0$ .
Find the particular antiderivative satisfying $F' (x) = cos 2 x$ and $F (0) = 3$ .
Find $\int x 1 + x^{2} d x$ .
Find $\int sin 2 x cos 5 x d x$ .
Find $\int d x / (x^{2} + 4 x + 8)$ .
Find $\int (3 x + 1) / 9 - 4 x - x^{2} d x$ on an appropriate interval.
Find $\int x / [(x - 1) (x + 2)] d x$ .
Find $\int d x / [(x + 1)^{2} (x + 2)]$ .
Find $\int (x^{3} + 1) / (x^{2} + 1) d x$ .
Find $\int x ln x d x$ for $x > 0$ .
Find $\int e^{x} (sin x + cos x) d x$ by recognising a product derivative.
Find $\int x^{2} + 2 x + 2 d x$ .
Evaluate $\int_{0}^{1} x e^{x} d x$ .
Evaluate $\int_{1}^{3} d x / [x^{2} (x + 1)]$ .
Evaluate $\int_{0}^{1} x / 1 + x^{2} d x$ by changing both limits.
Evaluate $\int_{- 1}^{1} x^{17} cos^{4} x d x$ and justify the method.
Evaluate $\int_{0}^{π /2} sin^{3} x d x$ .
Evaluate $\int_{0}^{π /4} 2 tan^{3} x d x$ .
Evaluate $\int_{0}^{1} sin^{- 1} x d x$ , interpreting the antiderivative's endpoint value by its limit.
Evaluate $\int_{1}^{4} [∣ x - 1∣ + ∣ x - 2∣ + ∣ x - 3∣] d x$ .
If $A (x) = \int_{0}^{x} 1 + t^{2} d t$ , find $A' (x)$ and $A (0)$ without evaluating the integral.
For a continuous function $g$ on $[a, b]$ satisfying $g (a + b - x) = g (x)$ , prove the weighted symmetry formula above and explain the midpoint factor.

Questions 13–20 have values $1$ , $2/3 + ln (2/3)$ , $2 - 1$ , $0$ , $2/3$ , $1 - ln 2$ , $π /2 - 1$ , and $19/2$ , respectively. Your solution should show the method, not only match these values.

Key Takeaways

Integration connects rates, antiderivatives, and accumulated values. Choose methods from the integrand's structure and combine them when necessary. Standard forms become visible through algebra, identities, shifts, and substitutions. The Fundamental Theorems link accumulation to differentiation and endpoint evaluation. Symmetry and sign analysis organise definite integrals before calculation. A complete answer includes correct reasoning, notation, domains, constants, and appropriate verification.

Previous · Lesson 8

Properties and Symmetry of Definite Integrals

End of chapter

One chapter, two questions, several connected tools

Choose a method by looking for evidence

Reconnect the standard formulas

Mixed indefinite integrals

Mixed definite integrals

Checking a solution from beginning to end

Quiz

Practice Problems

Key Takeaways