Lecture 28: Indefinite integrals

Common Question Types and Typical Examples in This Chapter

Exam Content

• (1) The concept and properties of indefinite integrals
• (2) Basic formulas for indefinite integrals
• (3) Three main integration methods
• (4) Integrals of three common integrable functions

Frequently Tested Question Types

• Finding indefinite integrals (substitution, division)

1.1 What is an indefinite integral?

Knowledge Points Distribution

• 2+3+3
• 2: Two concepts -> 1. antiderivative; 2. Indefinite integral;
• 3: Three methods: two types of substitution + distribution;
• 3: Three types of common integrals

Key Point: Three methods -> Two types of substitution + distribution;

Scope: Master the basic methods for evaluating indefinite integrals.

1.1.1 Basic Concepts

antiderivative

• Concept:
• $F^{\prime}(x)=f(x)$
• $G^{\prime}(x)=f(x)$
• Therefore, $G(x)-F(x)=C$.

#####Definition: Variablepoints

description: is a function. The indefinite integral (or antiderivative) of $f(x)$ is a derivative function equal to $f(x)$. $F(x)$, that is, $F′(x) = f (x)$, or written as $[F (x)+c]^{\prime}=f (x)$

$\int f(x)\,dx=F(x)+C$.

Explanation

• Concept:
• An indefinite integral represents the full family of antiderivatives of $f(x)$.
• Example:
• For example, in $\int f_{(x)}dx=F_{(x)}+C$:
• Let $F(x)$ be one antiderivative of $f(x)$.
• Call all antiderivative $F(x)+C$ of function $f(x)$ the indefinite integral of function $f(x)$, also called the antiderivative of function $f(x)$;
• Symbol:
• $\int$: Integral sign;
• $f(x)$: integrand;
• $dx$: Integral variable;
• $f(x)\,dx$: the integrand together with the differential
• $C$: constant of integration;

Geometric Meaning of Indefinite Integrals

• A family of curves that differ only by a vertical translation.

1.1.2 Existence of antiderivatives

#####Theorem: Theoriginalfunctionexistsbytheorem

description:

1. If $f(x)$ is continuous on interval $I$, then $f(x)$ must have an antiderivative on interval $I$;

2. If $f(x)$ has the first type of discontinuity point on the interval $I$, then $f(x)$ has no antiderivative on the interval $I$;

Explanation

• Relationships:
• For continuous functions, there must be an antiderivative;
• Continuous -> exists with an antiderivative;
• There exists an antiderivative -X> continuous;
• Conclusion:
• A discontinuous function may still have an antiderivative, but derivatives satisfy the intermediate value property and cannot have discontinuities of the first kind.

Examples

• $g(x)=\operatorname{sgn}x=\begin{cases}-1, &x<0, \\0, &x=0, \\1, &x>0.&\end{cases}$ has jump discontinuities -> belongs to the first type of discontinuities -> has no antiderivative;
• The natural candidate $|x|+C$ is not differentiable at $x=0$.

1.1.3 Basic Properties of Indefinite Integrals

Nature 1:

• $(\int f(x)\mathrm{d}x)^{\prime}=f(x)$
• $\mathrm{d}\int f(x)\mathrm{d}x=f(x)\mathrm{d}x$

Property 2:

• $\int f^{\prime}(x)\operatorname{d}x=f(x)+C$
• $\int\operatorname{d}f(x)=f(x)+C$

Nature 3:

• $\int[f(x)\pm g(x)]\operatorname{d}x=\int f(x)\operatorname{d}x\pm\int g(x)\operatorname{d}x$

Property 4:

• $\int kf(x)\operatorname{d}x=k\int f(x)\operatorname{d}x$

1.2 Common Integral Formulas

1.2.1 Integral Formula

$1、\int adx=ax+C\:,\:a$ is a constant

$2、\int x^{a}dx=\frac{x^{a+1}}{a+1}+C$, where a is a constant and $a\neq-1$

$3、\int\frac{1}{x}dx=\ln|x|+C$

$4、\int e^{x}dx=e^{x}+C$

$5、\int a^{x}dx=\frac{1}{\ln a}a^{x}+C$, where $a>0$ and $a\neq1$.

$6、\int\sin xdx=-\cos x+C$

7、$\int\cos xdx=\sin x+C$

$8、\int\sec^{2}xdx=\tan x+C$

$9、\int\csc^{2}xdx=-\cot x+C$

$10、\int\tan xdx=-\ln|\cos x|+C$

$11、\int\cot xdx=\ln\lvert\sin x\rvert+C$

$12、\int\sec xdx=\ln\lvert\sec x+\tan x\rvert+C$

$13、\int\csc xdx=-\ln\lvert\csc x+\cot x\rvert+C$

$14、\int\frac{dx}{1+x^2}=\arctan x+C$

$15、\int\frac1{\sqrt{a^2-x^2}}dx=\arcsin\frac xa+C$

$16、\int\frac1{x^2-a^2}dx=\frac1{2a}\ln\left|\frac{x-a}{x+a}\right|+C$

$17、\int\frac{dx}{\sqrt{1-x^2}}=\arcsin x+C$

$18、 \int\frac 1{a^2+x^2}dx=\frac 1 a\arctan\frac xa+C$

$19、\int\frac{dx}{\sqrt{x^2+a^2}}=\ln (x+\sqrt{x^2+a^2})+C$

$20、\int\frac{dx}{\sqrt{x^2-a^2}}=\ln\left|x+\sqrt{x^2-a^2}\right.|+C$

Common "Unable to Accumulate" Functions

• $\begin{aligned}&\int\mathrm{e}^{x^2}\mathrm{d}x\\&\int\frac{\sin x}x\mathrm{d}x\\&\int\frac{\cos x}x\mathrm{d}x\end{aligned}$

1.2.2 Trigonometric Function Summary

Basics of Trigonometric Functions

• Reciprocal relationship:
• $$\tan\alpha\cot\alpha=1、\sin\alpha\csc\alpha=1、\cos\alpha\sec\alpha=1$$
• Quotient relationship:
• $$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}、\cot\alpha=\frac{\cos\alpha}{\sin\alpha}$$
• Square relation:
• $$ \sin^{2}\alpha+\cos^{2}\alpha=1、1+\tan^{2}\alpha=\sec^{2}\alpha、1+\cot^{2}\alpha=\csc^{2}\alpha$$
• Double angle formula:
• $$\sin2\alpha=2\sin\alpha\cos\alpha$$
• $$\cos2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha$$
• $$\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{2\cot\alpha}{\cot^2\alpha-1}=\frac{2}{\cot\alpha-\tan\alpha}$$
• Lowering formula:
• $$\cos^{2}\alpha=\frac{1+\cos2\alpha}{2},\quad\sin^{2}\alpha=\frac{1-\cos2\alpha}{2},\quad\tan^2\alpha=\frac{1-\cos2\alpha}{1+\cos2\alpha}$$
• Inverse trigonometric function:
• If $x=a\sin t$, then $t=\arcsin\frac{x}{a}$.
• If $x=a\sec t$, then $t=\arccos\frac{a}{x}$.

Collection of Trigonometric Differentiations

• Sine and Cosine:
• $$(\sin x)^{\prime}=\cos x,\qquad(\cos x)^{\prime}=-\sin x$$
• tan、cot、sec、csc：
• $$\begin{aligned}(\tan x)^{\prime}&=\sec^2x&(\cot x)^{\prime}&=-\csc^2x\\\\(\sec x)^{\prime}&=\sec x\tan x&(\csc x)^{\prime}&=-\csc x\cot x\end{aligned}$$
• Inverse trigonometric function:
• $$\begin{aligned}&(\arcsin x)^{\prime}=\frac1{\sqrt{1-x^2}}&&(\arccos x)^{\prime}=-\frac1{\sqrt{1-x^2}}\\&(\arctan x)^{\prime}=\frac1{1+x^2}&&(\operatorname{arccot}x)^{\prime}=-\frac1{1+x^2}\end{aligned}$$

Common Trigonometric Integrals

• Corresponding one-to-one with differentiation:
• $$\int\sin xdx=-\cos x+C、\int\cos xdx=\sin x+C$$
• $$\int\tan xdx=-\ln|\cos x|+C、\int\cot xdx=\ln\lvert\sin x\rvert+C$$
• $$\int\sec xdx=\ln\lvert\sec x+\tan x\rvert+C、\int\csc xdx=-\ln\lvert\csc x+\cot x\rvert+C$$
• $$\int\sec^{2}xdx=\tan x+C、\int\csc^{2}xdx=-\cot x+C$$
• $$\int\tan x secx=\sec x+C、\int{\cot x\,\csc x}\,dx=-csc x+C$$

Common Inverse Trigonometric Integrals

• $$\int\frac{dx}{\sqrt{1-x^2}}=\arcsin x+C$$
• $$\int\frac1{\sqrt{a^2-x^2}}dx=\arcsin\frac xa+C$$
• $$\int\frac{dx}{1+x^2}=\arctan x+C$$
• $$\int\frac 1{a^2+x^2}dx=\frac 1 a\arctan\frac xa+C$$

1.3 Three Common Types of Integrable Functions

1. Rational Function Integral

• Points: $\int R(x)\operatorname{d}x$
• Rational function: $R(x)$
• (1) General method (partial fraction method);
• (2) Special method (additive and subtractive terms + square differential square)
• Examples of adding, detracting, and splitting items:
• $\int\frac1{t^2-1}\cdot\frac1{t+1}\mathrm{d}t.$
• Get: $\int\frac1{(t^2-1)(t+1)}\mathrm{d}t=\frac12\int\frac{(t+1)-(t-1)}{(t^2-1)(t+1)}\mathrm{d}t=\frac14\ln\left|\frac{t-1}{t+1}\right|+\frac1{2(t+1)}+C$

2. Trigonometric Rational Integral

• Points: $\int R(\sin x,\cos x)\mathrm{d}x$
• Explanation: Represents $\sin x$ and $\cos x$ obtained through rational operations;
• General method: universal replacement
• Ling $\tan\frac x2=t$
• $\int R(\sin x,\cos x)\operatorname{d}x=\int R(\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2})\frac2{1+t^2}dt$
• Special methods (triangle deformation, substitution, division)
• 1. $\text{ If }R(-\sin x,\cos x)=-R(\sin x,\cos x)$, then let $u=\cos x$
• 2. $\text{ If }R(\sin x,-\cos x)=-R(\sin x,\cos x)$, then let $u=\sin x$
• 3. $\text{ If }R(-\sin x,-\cos x)=R(\sin x,\cos x)$, then let $u=\tan x$
• Example:
• Special method: $\int\frac{\operatorname{d}x}{1+\sin x}=\int\frac{1-\sin x}{\cos^2x}\mathrm{d}x=\tan x-\frac1{\cos x}+C.$
• General method: Let $\tan\frac x2=t$, then $\begin{aligned}\text{ Original form }& =\int\frac 1{1+\frac{2 t}{1+t^2}}\cdot{\frac 2{1+t^2}dt} \\&=\int\frac{2 dt}{\left (1+t\right)^2}=-\frac 2{1+t}+C\end{aligned}$

3. Simple Irrational Integral

• Scoring Method:
• $$\int R(x,\sqrt[n]{\frac{ax+b}{cx+d}})\operatorname{d}x$$
• Explanation:
• Example: In the two functions of $\int\frac{dx}{(2-x)\sqrt{1-x}}$:
• 1. General function part: $(2-x)$
• 2. Radical function: $\sqrt{1-x}$
• General Method:
• $$\text{ Order }\sqrt[n]{\frac{ax+b}{cx+d}}=t$$
• Example:
• $\int\frac1x\sqrt{\frac{x+1}x}dx$
• $\text{ Order }\sqrt{\frac{x+1}x}=t\text{ , then }\quad x=\frac1{t^2-1},dx=-\frac{2t}{\left(t^2-1\right)^2}dt,\int\frac1x\sqrt{\frac{x+1}x}dx=\int(t^2-1)t\frac{-2t}{(t^2-1)^2}dt=-2\int\left(1+\frac1{t^2-1}\right)dt$

1.4 Frequently Tested Question Types

Question Type: Findthereplacementoftheindefiniteintegral. Sections

PART 1: Problem-solving methods

About the Two Types of Substitution Methods

• Many problems can be solved using the first type of substitution method and the derivative method of multiplication, as well as the second type of substitution method and the t-composite function;
• When the first type of substitution method is available, the first type of substitution method is prioritized;

On Piecewise Functions for Calculating Indefinite Integrals

• Note: At point 0, whether the segmental function can be differentiated and observe its continuity;
• Discontinuous -> non-differentiable -> incorrect antiderivative;
• Method:
• 1. For piecewise functions to find indefinite integrals, first find the indefinite integral normally, then write $C1、C2$ when writing C;
• 2. Solve thewith C resultsof the two functions at the segmentation points;
• 3. According to the requirement of continuous boundary points, the two functions with C must have the same result, so find the relationship between C1 and C2;
• 4. Bring C1, C 2 into antiderivative in the form of C;
• Supplement: an implicit premise
• As long as the integrand is a piecewise continuous function, as long as continuity is ensured, differentiability can also be guaranteed;

Multiplying Two Different Functions

• When multiplying two different types of functions, theDistributed Integral Methodcan be considered;
• Incorporate the easy-to-get parts into the DX;

Given the antiderivative problem

• When the question is a known antiderivative, find the content of $f(x)$ in the indefinite integral, and there is a derivative in the indefinite integral, you can directly extract the derivative into dx, thus directly bringing the differentiated result of the antiderivative;

PART 2: Typical Example Problems

Example Question: $\text{ Let }f(x)=\begin{cases}&e^x,&x\geq0,\\&\cos x,&x<0,&\end{cases}\text{ then }\int f(x)dx=$

• Analysis
• $\left.\int f(x)dx=\left\{\begin{array}{l}{{e^{x}+c_{1}},}&{{{x\geq0}}}\\{{\sin^{\prime}x+c_{2}^{\prime}}}&{{x<0}}\\\end{array}\right.\right.$
• At this point, as x tends toward positive 0 and negative 0 respectively, we get: $1+C_1=C_2$
• Therefore, let $C_1=C$, then $C_2=1+C$, substitute it into the original form;
• $\left.\int f(x)dx=\left\{\begin{array}{ll}{{e^{x}+c,\quad x\geq0}}\\{{sinx+1+c.\quad x<0}}\\\end{array}\right.\right.$
• Analysis
• Question Type: #

Example Question: $\text{ Calculation }\int\frac{x^2}{\sqrt{a^2-x^2}}dx(a>0).$

• Analysis
• Summary: When $x=a\sin t$, $\sin t\cos t=\frac{x}{a}\sqrt{1-\frac{x^{2}}{a^{2}}}$.
• Where $\frac{x}{a}$ is $\sin t$, $\sqrt{1-\frac{x^{2}}{a^{2}}}$ is given by $\sin^{2}t+\cos^{2}t=1$.
• Analysis
• Question Type: #

Example: Given $\frac{\sin x}x$ is an antiderivative of f(x), find $\int x^3f^{\prime}(x)\operatorname{d}x$

• Analysis
• Directly set $\int x^{3}f'(x)\mathsf{d}x=\int x^{3}df(x)$ and then directly divide into parts;
• $=x^3f(x)-3\int x^2\mathrm{d}\left(\frac{\sin x}x\right)$
• Analysis
• Question Type: #

PART 3: Key Points Review

Summary:

• When $x=a\sin t$, $\sin t\cos t=\frac{x}{a}\sqrt{1-\frac{x^{2}}{a^{2}}}$.
• Where $\frac{x}{a}$ is $\sin t$, $\sqrt{1-\frac{x^{2}}{a^{2}}}$ is given by $\sin^{2}t+\cos^{2}t=1$.