4 Indefinite Integrals

4.1 The Concept of Indefinite Integrals

A function $F$ is called an antiderivative of a function $f$ on an interval $I \subset \mathbb{R}$ if

$$\begin{equation*} F'(x) = f(x), \qquad \forall x \in I. \end{equation*}$$

The indefinite integral of $f$ on $I \subset \mathbb{R}$ is the family of all antiderivatives of $f$ on $I$, and is denoted by

$$\begin{equation*} \int f(x)\, dx. \end{equation*}$$

Let $I \subset \mathbb{R}$ be an interval. If $F_1$ and $F_2$ are two antiderivatives of $f$ on $I$, then the difference $F_1(x) - F_2(x)$ is a constant. Consequently,

$$\begin{equation*} \int f(x)\, dx = F(x) + C, \end{equation*}$$

where $F$ is any antiderivative of $f$ on $I$, and $C$ is an arbitrary constant.

Examples

1. $$\begin{equation*} \int e^{x}\, dx = e^{x} + C. \end{equation*}$$
2. $$\begin{equation*} \int \sin x\, dx = -\cos x + C, \end{equation*}$$

$$\begin{equation*} \int \cos x\, dx = \sin x + C. \end{equation*}$$

$$\begin{equation*} \int \sinh x\, dx = \cosh x + C, \end{equation*}$$ $$\begin{equation*} \int \cosh x\, dx = \sinh x + C. \end{equation*}$$

Where

$$\begin{equation*} \sinh x = \frac{e^{x} - e^{-x}}{2},\qquad \cosh x = \frac{e^{x} + e^{-x}}{2}. \end{equation*}$$

4. For $\alpha \ne -1$,

   $$\begin{equation*} \int x^{\alpha}\, dx = \frac{x^{\alpha+1}}{\alpha+1} + C. \end{equation*}$$
5. $$\begin{equation*} \int \frac{1}{x}\, dx = \ln|x| + C. \end{equation*}$$
6. $$\begin{equation*} \int \frac{1}{1+x^{2}}\, dx = \arctan x + C. \end{equation*}$$

4.2 The Computation of Indefinite Integrals

Theorem 4.1. If $F$ and $G$ are antiderivatives of $f$ and $g$ on an interval $I$, then $\lambda F + \mu G$ is an antiderivative of $\lambda f + \mu g$ on $I$. Consequently,

$$\begin{equation*} \int \bigl(\lambda f(x) + \mu g(x)\bigr)\, dx = \lambda \int f(x)\, dx + \mu \int g(x)\, dx. \end{equation*}$$

Theorem 4.2. If $F$ and $G$ are antiderivatives of $f$ and $g$, respectively, then

$$\begin{equation*} \int f(G(x))\, g(x)\, dx = F(G(x)) + C. \end{equation*}$$

Theorem 4.3. If $F$ and $G$ are antiderivatives of $f$ and $g$ on an interval $I$, then

$$\begin{equation*} \int F(x)\, g(x)\, dx = F(x) G(x) - \int G(x)\, f(x)\, dx. \end{equation*}$$

Examples

1. Compute

   $$\begin{equation*} \int \frac{x}{1+x^{2}}\, dx. \end{equation*}$$

   Let

   $$\begin{equation*} y = 1 + x^{2}, \qquad dy = 2x\, dx. \end{equation*}$$

   Then

   $$\begin{equation*} \int \frac{x}{1+x^{2}}\, dx = \frac{1}{2} \int \frac{1}{1+x^{2}}\, d(1+x^{2}) = \frac{1}{2} \int \frac{1}{y}\, dy. \end{equation*}$$

   Thus

   $$\begin{equation*} \frac{1}{2}\int \frac{1}{y}\, dy = \frac{1}{2}\ln|y| + C = \frac{1}{2}\ln(1+x^{2}) + C. \end{equation*}$$

2. Compute

$$\begin{equation*} \int \frac{1}{\cos^{2m} x}\, dx. \end{equation*}$$

Let

$$\begin{equation*} t = \tan x, \qquad dt = \sec^{2}x\, dx = \frac{1}{\cos^{2}x}\, dx. \end{equation*}$$

Then

$$\begin{equation*} \int \frac{1}{\cos^{2m} x}\, dx = \int \frac{1}{\cos^{2m-2} x}\, d(\tan x) = \int (1+t^{2})^{\,m-1}\, dt. \end{equation*}$$

Thus the integral can be reduced to a polynomial-type integral in $t$:

$$\begin{equation*} \int \frac{1}{\cos^{2m} x}\, dx = \int (1+t^{2})^{\,m-1}\, dt, \qquad t=\tan x. \end{equation*}$$

3. Compute

$$\begin{equation*} \int f(\cos^{2} x)\, dx. \end{equation*}$$

We first rewrite the integrand:

$$\begin{equation*} \int f(\cos^{2} x)\, dx = \int \frac{f(\cos^{2} x)\,\cos^{2}x}{\cos^{2}x}\, dx. \end{equation*}$$

Let

$$\begin{equation*} t = \tan x, \qquad dt = \frac{1}{\cos^{2}x}\, dx, \qquad \cos^{2}x = \frac{1}{1+t^{2}}. \end{equation*}$$

Then

$$\begin{equation*} \int f(\cos^{2} x)\, dx = \int \frac{1}{1+t^{2}}\, f\!\left(\frac{1}{1+t^{2}}\right)\, dt. \end{equation*}$$

Thus the integral of $f(\cos^{2}x)$ is transformed into an integral in $t$:

$$\begin{equation*} \int f(\cos^{2} x)\, dx = \int \frac{1}{1+t^{2}}\, f\!\left(\frac{1}{1+t^{2}}\right)\, dt, \qquad t=\tan x. \end{equation*}$$

4. Compute

   $$\begin{equation*} \int \frac{1}{\sqrt{x^{2}+a^{2}}}\, dx \end{equation*}$$

   Let

   $$\begin{equation*} \theta = \arctan\!\left(\frac{x}{a}\right) \end{equation*}$$

   Then

   $$\begin{equation*} \int \frac{1}{\sqrt{x^{2}+a^{2}}}\, dx = \int \frac{1}{\sqrt{a^{2}\tan^{2}\theta + a^{2}}}\, d(a\tan\theta) \end{equation*}$$

   Since

   $$\begin{equation*} \sqrt{a^{2}\tan^{2}\theta + a^{2}} = a\sqrt{1+\tan^{2}\theta} = \frac{a}{\cos\theta}, \end{equation*}$$

   we have

   $$\begin{equation*} \int \frac{1}{\sqrt{x^{2}+a^{2}}}\, dx = \int \frac{1}{\cos\theta}\, d\theta \end{equation*}$$

   Let

   $$\begin{equation*} t = \sin\theta, \qquad dt = \cos\theta\, d\theta . \end{equation*}$$

   Then

   $$\begin{equation*} \int \frac{1}{\cos\theta}\, d\theta = \frac{1}{2}\int \left(\frac{1}{1-t} + \frac{1}{1+t}\right) dt = \frac{1}{2}\ln\frac{1+t}{1-t} + C \end{equation*}$$

   Thus

   $$\begin{equation*} \frac{1}{2}\ln\frac{1+\sin\theta}{1-\sin\theta} + C \end{equation*}$$

   Since

   $$\begin{equation*} \sin\theta = \frac{x}{\sqrt{a^{2}+x^{2}}}, \end{equation*}$$

   we get

   $$\begin{equation*} \frac{1}{2}\ln \left( \frac{1+\frac{x}{\sqrt{a^{2}+x^{2}}}} {1-\frac{x}{\sqrt{a^{2}+x^{2}}}} \right) = \ln\left(x+\sqrt{a^{2}+x^{2}}\right) + C_1. \end{equation*}$$

   Therefore

   $$\begin{equation*} \int \frac{1}{\sqrt{x^{2}+a^{2}}}\, dx = \ln\left(x+\sqrt{a^{2}+x^{2}}\right) + C . \end{equation*}$$

5. Let

   $$\begin{equation*} I_n = \int e^{x}\sin^{n}x\, dx . \end{equation*}$$

   Then

   $$\begin{equation*} I_n = \int e^{x}\sin^{n}x\, dx = \int \sin^{n}x\, d(e^{x}) = e^{x}\sin^{n}x - \int e^{x} d(\sin^{n}x). \end{equation*}$$

   Compute the differential:

   $$\begin{equation*} d(\sin^{n}x) = n\sin^{\,n-1}x \cos x\, dx. \end{equation*}$$

   Hence

   $$\begin{equation*} I_n = e^{x}\sin^{n}x - n\int \sin^{\,n-1}x \cos x\, e^{x}\, dx. \end{equation*}$$

   Rewrite the integral:

   $$\begin{equation*} I_n = e^{x}\sin^{n}x - n e^{x}\sin^{\,n-1}x\cos x + n\int e^{x} d(\sin^{\,n-1}x\cos x). \end{equation*}$$

   Thus

   $$\begin{equation*} I_n = e^{x}\bigl(\sin^{n}x - n\sin^{\,n-1}x\cos x\bigr) + n(n-1)(I_{n-2} - I_n) - n I_n . \end{equation*}$$

   Solve for (I_n):

   $$\begin{equation*} I_n = \frac{ e^{x}\sin^{\,n-1}x\bigl(\sin x - n\cos x\bigr) + n(n-1)I_{n-2} }{n^{2}+1}. \end{equation*}$$

   Special case:

   $$\begin{equation*} I_1 = \frac{e^{x}(\sin x - \cos x)}{2}. \end{equation*}$$

6. A rational function is a quotient of two polynomials:

   $$\begin{equation*} R(x)=\frac{P(x)}{Q(x)}, \qquad P, Q \in \mathbb{R}[x], \quad Q(x)\neq 0. \end{equation*}$$

   To integrate $R(x)$, follow these steps.

​ Polynomial Long Division

​ If

$$\begin{equation*} \deg P \ge \deg Q, \end{equation*}$$

​ divide $P$ by $Q$:

$$\begin{equation*} \frac{P(x)}{Q(x)} = S(x) + \frac{A(x)}{Q(x)}, \qquad \deg A < \deg Q. \end{equation*}$$

​ Thus,

$$\begin{equation*} \int R(x)\,dx = \int S(x)\,dx + \int \frac{A(x)}{Q(x)}\, dx. \end{equation*}$$

​ Factor the Denominator

​ Factor $Q(x)$ into irreducible real factors:

$$\begin{equation*} Q(x)= \prod (x-a_i)^{k_i} \prod (x^2 + b_j x + c_j)^{m_j}, \end{equation*}$$

​ where each quadratic satisfies

$$\begin{equation*} b_j^2 - 4c_j < 0. \end{equation*}$$

​

​ Partial Fraction Decomposition

$$\begin{equation*} \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_k}{(x-a)^k} \end{equation*}$$ $$\begin{equation*} \frac{A_1x+B_1}{x^2+bx+c} + \frac{A_2x+B_2}{(x^2+bx+c)^2} + \cdots + \frac{A_mx+B_m}{(x^2+bx+c)^m}. \end{equation*}$$

​ Integrate Each Term

$$\begin{equation*} \int \frac{1}{x-a}\, dx = \ln |x-a| + C. \end{equation*}$$ $$\begin{equation*} \int \frac{1}{(x-a)^k}\, dx = \frac{-(x-a)^{-k+1}}{k-1} + C \qquad (k\ge 2). \end{equation*}$$ $$\begin{equation*} x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + d, \quad d = c-\frac{b^2}{4}. \end{equation*}$$ $$\begin{equation*} \int \frac{1}{(x+\frac{b}{2})^2+d}\, dx = \frac{1}{\sqrt{d}} \arctan\!\left(\frac{x+\frac{b}{2}}{\sqrt{d}}\right) + C. \end{equation*}$$ $$\begin{equation*} \int \frac{x+\frac{b}{2}}{(x+\frac{b}{2})^2+d}\, dx = \frac{1}{2}\ln\left((x+\frac{b}{2})^2+d\right) + C. \end{equation*}$$

​

6. $$\begin{equation*} I_n = \int \frac{1}{(1+x^2)^n}\,dx \qquad (n\ge 2). \end{equation*}$$

   Write

   $$\begin{equation*} I_n = \int \frac{1}{(1+x^2)^n}\,dx = \int \frac{(1+x^2)-x^2}{(1+x^2)^n}\,dx = I_{n - 1} - \int \frac{x^2}{(1+x^2)^n}\,dx. \end{equation*}$$ $$\begin{equation*} \begin{aligned} &\int \frac{x^2}{(1+x^2)^n}\,dx = \frac{1}{2}\int \frac{x}{(1+x^2)^{n}} \,d(1 + x^2) = \frac{1}{2 (1 - n)}\int x \,d\left((1 + x^2) ^ {1 - n}\right) \\ &= \frac{1}{2 (1 - n)} \left( \frac{x}{(1+x^2)^{n-1}} - \int \frac{1}{(1 + x^2)^{n - 1}} \, dx\right). \end{aligned} \end{equation*}$$

   Therefore

   $$\begin{equation*} I_n = \left(1 - \frac{1}{2(n-1)}\right) I_{n-1} + \frac{1}{2(n-1)}\frac{x}{(1+x^2)^{n-1}}. \end{equation*}$$

   Thus the reduction formula is

   $$\begin{equation*} I_n = \frac{x}{2(n-1)(1+x^2)^{n-1}} + \frac{2n-3}{2(n-1)}\, I_{n-1}, \qquad n\ge 2. \end{equation*}$$

   Together with

   $$\begin{equation*} I_1 = \int \frac{dx}{1+x^2} = \arctan x + C, \end{equation*}$$
7. $$\begin{equation*} \int R(\cos\theta,\sin\theta)\, d\theta = \int R\!\left( \frac{1 - t^{2}}{1 + t^{2}},\; \frac{2t}{1 + t^{2}} \right) \frac{2}{1+t^{2}}\, dt. \end{equation*}$$

$$\begin{equation*} \int R(x,\sqrt{x^{2}-1})\, dx = \int R\!\left( \frac{1+t^{2}}{1-t^{2}},\; \frac{2t}{\,1-t^{2}\,} \right) \, d\!\left(\frac{1+t^{2}}{1-t^{2}}\right). \end{equation*}$$