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 \[ F'(x) = f(x), \qquad \forall x \in I. \]

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

\[ \int f(x)\, dx. \]

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, \[ \int f(x)\, dx = F(x) + C, \]

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

Examples

1. \[ \int e^{x}\, dx = e^{x} + C. \]

2. \[ \int \sin x\, dx = -\cos x + C, \]

\[ \int \cos x\, dx = \sin x + C. \]

\[ \int \sinh x\, dx = \cosh x + C, \]

\[ \int \cosh x\, dx = \sinh x + C. \]

Where \[ \sinh x = \frac{e^{x} - e^{-x}}{2},\qquad \cosh x = \frac{e^{x} + e^{-x}}{2}. \]

4. For \(\alpha \ne -1\), \[ \int x^{\alpha}\, dx = \frac{x^{\alpha+1}}{\alpha+1} + C. \]

5. \[ \int \frac{1}{x}\, dx = \ln|x| + C. \]

6. \[ \int \frac{1}{1+x^{2}}\, dx = \arctan x + C. \]

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, \[ \int \bigl(\lambda f(x) + \mu g(x)\bigr)\, dx = \lambda \int f(x)\, dx + \mu \int g(x)\, dx. \]

Theorem 4.2 If \(F\) and \(G\) are antiderivatives of \(f\) and \(g\), respectively, then \[ \int f(G(x))\, g(x)\, dx = F(G(x)) + C. \]

Theorem 4.3 If \(F\) and \(G\) are antiderivatives of \(f\) and \(g\) on an interval \(I\), then \[ \int F(x)\, g(x)\, dx = F(x) G(x) - \int G(x)\, f(x)\, dx. \]

Examples

1. Compute \[ \int \frac{x}{1+x^{2}}\, dx. \]

   Let \[ y = 1 + x^{2}, \qquad dy = 2x\, dx. \]

   Then \[ \int \frac{x}{1+x^{2}}\, dx = \frac{1}{2} \int \frac{1}{1+x^{2}}\, d(1+x^{2}) = \frac{1}{2} \int \frac{1}{y}\, dy. \]

   Thus \[ \frac{1}{2}\int \frac{1}{y}\, dy = \frac{1}{2}\ln|y| + C = \frac{1}{2}\ln(1+x^{2}) + C. \]

2. Compute \[ \int \frac{1}{\cos^{2m} x}\, dx. \]

Let \[ t = \tan x, \qquad dt = \sec^{2}x\, dx = \frac{1}{\cos^{2}x}\, dx. \]

Then \[ \int \frac{1}{\cos^{2m} x}\, dx = \int \frac{1}{\cos^{2m-2} x}\, d(\tan x) = \int (1+t^{2})^{\,m-1}\, dt. \]

Thus the integral can be reduced to a polynomial-type integral in \(t\): \[ \int \frac{1}{\cos^{2m} x}\, dx = \int (1+t^{2})^{\,m-1}\, dt, \qquad t=\tan x. \]

3. Compute \[ \int f(\cos^{2} x)\, dx. \]

We first rewrite the integrand: \[ \int f(\cos^{2} x)\, dx = \int \frac{f(\cos^{2} x)\,\cos^{2}x}{\cos^{2}x}\, dx. \]

Let \[ t = \tan x, \qquad dt = \frac{1}{\cos^{2}x}\, dx, \qquad \cos^{2}x = \frac{1}{1+t^{2}}. \]

Then \[ \int f(\cos^{2} x)\, dx = \int \frac{1}{1+t^{2}}\, f\!\left(\frac{1}{1+t^{2}}\right)\, dt. \]

Thus the integral of \(f(\cos^{2}x)\) is transformed into an integral in \(t\): \[ \int f(\cos^{2} x)\, dx = \int \frac{1}{1+t^{2}}\, f\!\left(\frac{1}{1+t^{2}}\right)\, dt, \qquad t=\tan x. \]

4. Compute \[ \int \frac{1}{\sqrt{x^{2}+a^{2}}}\, dx \]

   Let \[ \theta = \arctan\!\left(\frac{x}{a}\right) \]

   Then \[ \int \frac{1}{\sqrt{x^{2}+a^{2}}}\, dx = \int \frac{1}{\sqrt{a^{2}\tan^{2}\theta + a^{2}}}\, d(a\tan\theta) \]

   Since \[ \sqrt{a^{2}\tan^{2}\theta + a^{2}} = a\sqrt{1+\tan^{2}\theta} = \frac{a}{\cos\theta}, \]

   we have \[ \int \frac{1}{\sqrt{x^{2}+a^{2}}}\, dx = \int \frac{1}{\cos\theta}\, d\theta \]

   Let \[ t = \sin\theta, \qquad dt = \cos\theta\, d\theta . \]

   Then \[ \int \frac{1}{\cos\theta}\, d\theta = \frac{1}{2}\int \left(\frac{1}{1-t} + \frac{1}{1+t}\right) dt = \frac{1}{2}\ln\frac{1+t}{1-t} + C \]

   Thus \[ \frac{1}{2}\ln\frac{1+\sin\theta}{1-\sin\theta} + C \]

   Since \[ \sin\theta = \frac{x}{\sqrt{a^{2}+x^{2}}}, \]

   we get \[ \frac{1}{2}\ln \left( \frac{1+\frac{x}{\sqrt{a^{2}+x^{2}}}} {1-\frac{x}{\sqrt{a^{2}+x^{2}}}} \right) = \ln\left(x+\sqrt{a^{2}+x^{2}}\right) + C_1. \]

   Therefore \[ \int \frac{1}{\sqrt{x^{2}+a^{2}}}\, dx = \ln\left(x+\sqrt{a^{2}+x^{2}}\right) + C . \]

5. Let \[ I_n = \int e^{x}\sin^{n}x\, dx . \]

   Then \[ I_n = \int e^{x}\sin^{n}x\, dx = \int \sin^{n}x\, d(e^{x}) = e^{x}\sin^{n}x - \int e^{x} d(\sin^{n}x). \]

   Compute the differential: \[ d(\sin^{n}x) = n\sin^{\,n-1}x \cos x\, dx. \]

   Hence \[ I_n = e^{x}\sin^{n}x - n\int \sin^{\,n-1}x \cos x\, e^{x}\, dx. \]

   Rewrite the integral: \[ I_n = e^{x}\sin^{n}x - n e^{x}\sin^{\,n-1}x\cos x + n\int e^{x} d(\sin^{\,n-1}x\cos x). \]

   Thus \[ I_n = e^{x}\bigl(\sin^{n}x - n\sin^{\,n-1}x\cos x\bigr) + n(n-1)(I_{n-2} - I_n) - n I_n . \]

   Solve for \(I_n\): \[ I_n = \frac{ e^{x}\sin^{\,n-1}x\bigl(\sin x - n\cos x\bigr) + n(n-1)I_{n-2} }{n^{2}+1}. \]

   Special case: \[ I_1 = \frac{e^{x}(\sin x - \cos x)}{2}. \]

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

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

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

   Polynomial Long Division

   If \[ \deg P \ge \deg Q, \] divide \(P\) by \(Q\): \[ \frac{P(x)}{Q(x)} = S(x) + \frac{A(x)}{Q(x)}, \qquad \deg A < \deg Q. \] Thus, \[ \int R(x)\,dx = \int S(x)\,dx + \int \frac{A(x)}{Q(x)}\, dx. \]

   Factor the Denominator

   Factor \(Q(x)\) into irreducible real factors:

\[ Q(x)= \prod (x-a_i)^{k_i} \prod (x^2 + b_j x + c_j)^{m_j}, \]

where each quadratic satisfies \[ b_j^2 - 4c_j < 0. \]

Partial Fraction Decomposition \[ \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_k}{(x-a)^k} \]

\[ \frac{A_1x+B_1}{x^2+bx+c} + \frac{A_2x+B_2}{(x^2+bx+c)^2} + \cdots + \frac{A_mx+B_m}{(x^2+bx+c)^m}. \]

Integrate Each Term \[ \int \frac{1}{x-a}\, dx = \ln |x-a| + C. \]

\[ \int \frac{1}{(x-a)^k}\, dx = \frac{-(x-a)^{-k+1}}{k-1} + C \qquad (k\ge 2). \]

\[ x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + d, \quad d = c-\frac{b^2}{4}. \]

\[ \int \frac{1}{(x+\frac{b}{2})^2+d}\, dx = \frac{1}{\sqrt{d}} \arctan\!\left(\frac{x+\frac{b}{2}}{\sqrt{d}}\right) + C. \]

\[ \int \frac{x+\frac{b}{2}}{(x+\frac{b}{2})^2+d}\, dx = \frac{1}{2}\ln\left((x+\frac{b}{2})^2+d\right) + C. \]

6. \[ I_n = \int \frac{1}{(1+x^2)^n}\,dx \qquad (n\ge 2). \]

   Write

   \[ I_n = \int \frac{1}{(1+x^2)^n}\,dx = \int \frac{(1+x^2)-x^2}{(1+x^2)^n}\,dx = I_{n - 1} - \int \frac{x^2}{(1+x^2)^n}\,dx. \]

   \[ \begin{aligned} &\int \frac{x^2}{(1+x^2)^n}\,dx = \frac{1}{2}\int \frac{x}{(1+x^2)^{n}} \,d(1 + x^2) = \frac{1}{2 (1 - n)}\int x \,d\left((1 + x^2) ^ {1 - n}\right) \\ &= \frac{1}{2 (1 - n)} \left( \frac{x}{(1+x^2)^{n-1}} - \int \frac{1}{(1 + x^2)^{n - 1}} \, dx\right). \end{aligned} \]

   Therefore

   \[ I_n = \left(1 - \frac{1}{2(n-1)}\right) I_{n-1} + \frac{1}{2(n-1)}\frac{x}{(1+x^2)^{n-1}}. \]

   Thus the reduction formula is \[ I_n = \frac{x}{2(n-1)(1+x^2)^{n-1}} + \frac{2n-3}{2(n-1)}\, I_{n-1}, \qquad n\ge 2. \]

   Together with

   \[ I_1 = \int \frac{dx}{1+x^2} = \arctan x + C, \]

7. \[ \int R(\cos\theta,\sin\theta)\, d\theta = \int R\!\left( \frac{1 - t^{2}}{1 + t^{2}},\; \frac{2t}{1 + t^{2}} \right) \frac{2}{1+t^{2}}\, dt. \]

\[ \int R(x,\sqrt{x^{2}-1})\, dx = \int R\!\left( \frac{1+t^{2}}{1-t^{2}},\; \frac{2t}{\,1-t^{2}\,} \right) \, d\!\left(\frac{1+t^{2}}{1-t^{2}}\right). \]