Basics

This section introduces the concept of antidifferentiation, linearity, power rule and trigonometric identities.

Definition

Antiderivatives are the inverse of the differentiation. Given some function f(x) where

f^\prime(x) = g(x)

the antiderivative of g(x) is f(x). In general, the antiderivative of a function g(x) is a separate function f(x) whose we must differentiate to obtain g(x). This yields our definition.

Definition

The antiderivative of a function g(x) is some function f(x) such that

f^\prime(x) = g(x).

This is notated with an integration sign. Given a function g(x) we denote it’s antiderivative with

\int g(x) \, dx

These antiderivatives are also called indefinite integrals, or just integrals for short. For any function g(x), there exists an infinite set of antiderivatives f(x) which only vary by some constant. Thus we write

\int g(x) \, dx = f(x) + C

where f(x) is some antiderivative of g(x) and C is any real number. Antiderivatives do not always exists for any arbitrary function g(x), and when they do exist, they may not be possible to express in elementary terms.

From this point on, we will refer almost exclusively to antiderivatives as integrals. We will call functions integrable if their antiderivative exists.

Linearity

Integrals have a property called linearity. This means they are closed under linear transformations of functions like addition and scaler multiplication. For example

\int f(x) + g(x) \,dx = \int f(x) \,dx + \int g(x) \,dx.

Or, given some constant C

\int C f(x) \,dx = C \int f(x) \,dx.

This is presented in the following theorem without proof.

Theorem

For all integrable functions f(x),g(x) and constants \alpha,\beta \in \mathbb{R}, the following holds

\int \alpha f(x) + \beta g(x) \,dx = \alpha \int f(x) \,dx+ \beta \int g(x) \,dx

Inverse power rule

One of the basic rules of differentiation is that of power rule.

\frac{d}{dx} x^m = mx^{m-1}

Suspecting this will be a useful property, we can derive an expression for the inverse power rule from the definition of an integral.

\int mx^{m-1} dx = x^m

Apply linearity and the substituion n = m-1

\int x^{n} dx = \frac{x^{n+1}}{n+1}

Find the solution to the following integral

\int 2x^2 - 3x - 1 \, dx

We first begin by applying linearity

= 2 \int x^2 \,dx - 3\int x \, dx - \int 1x^0 \, dx

Then inverse power rule

= \frac{2x^3}{3} - \frac{3x^2}{2} - x + C

Giving us a solution to the polynomial. This process can be done for any standard polynomial.

Theorem (Integration of a polynomial)

Let f(x) be a polynomial in the form

f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + c

Then

\int f(x)\,dx = \frac{a_nx^{n+1}}{n+1} + \frac{a_{n-1}x^{n}}{n} + \cdots + \frac{a_1x^2}{2} + cx + C

Edge case

One may notice that when n=-1, inverse power rules gives us an undefined result

\int x^{-1}\,dx = \frac{1}{0} + C.

For this particular value of n, we get the expression

\int x^{-1}\,dx = \ln|x| + C.

Note the absolute value bars. This is because the function \ln is not defined for negative values of x, where our integrand is.

Common identities

This section will cover many commonly used integration identities without much explanation.

Trig identities

Trigonometry is a huge part of the integration bee, and will repeatedly appear in almost every section of this guide. As such it is crucial to have a complete understanding of these functions and how they interact.

The integrals below are the direct inverse of differentiating the six primary trig functions.

\begin{array}{ll} \displaystyle\int{{\cos x\,dx}} = \sin x + C \hspace{0.5in} & \displaystyle\int{{\sin x\,dx}} = - \cos x + C\\ \displaystyle\int{{{{\sec }^2}x\,dx}} = \tan x + C \hspace{0.5in} & \displaystyle\int{{{{\csc }^2}x\,dx}} = - \cot x + C\\ \displaystyle\int{{\sec x\tan x\,dx}} = \sec x + C \hspace{0.5in} & \displaystyle\int{{\csc x\cot x\,dx}} = - \csc x + C \end{array}

It is helpful (for me at least) to memorize the left column of the table, and then derive the right side by replacing each function with its ‘compliment’, and then negating the result.

Inverse Trig identities

We can use special formulas to differentiate inverse functions. By differentiating \arcsin(x) and \arctan(x), we get

\int \frac{1}{x^2 + 1} \, dx = \arctan(x) + C \hspace{0.5in} \int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin(x) + C

It is worth noting that

\frac{d}{dx}\arccos(x) = -\frac{1}{1-x^2}.

So the second integral can also be solved

\int \frac{1}{\sqrt{1-x^2}}\,dx = -\arccos(x) + C.

This is because \arcsin and -\arccos differ by a constant (\frac{\pi}{2}).

Exponential and logarithmic functions

By once again inverting common derivative rules, we get the following identities.

\int e^x \, dx = e^x + C \hspace{0.5in} \int a^x \,dx = \frac{a^x}{\ln{x}} + C

Examples

Some example problems from various websites

Inverse power rule

Pauls Online Notes (Calculus - 5.1.1)

Evaluate the following integral

\int 6x^5 - 18x^2 + 7 \, dx

Using the formula for evaluating polynomials

\int 6x^5 - 18x^2 + 7 \, dx = x^6 - 6x^3 + 7x + C

Pauls Online Notes (Calculus - 5.1.1)

Evaluate the following integral

\int 6x^5\, dx - 18x^2 + 7

Notice the integral ends at the dx, meaning the solution is

\int 6x^5 - 18x^2 + 7 \, dx = x^6 - 18x^2 + C

Pauls Online Notes (Calculus - 5.1.3)

Evaluate the following integral

\int 12t^7 - t^2 - t + 3 \, dt

Classic polynomial, the only difference is the use of t instead of x

\int 12t^7 - t^2 - t + 3 \, dt = \frac{3}{2}t^8 - \frac{1}{3}t^3 - \frac{1}{2}t^2 + 3t + C

Pauls Online Notes (Calculus - 5.2.6)

Evaluate the following integral

\int \sqrt[3]{w} + 10\sqrt[5]{w^3}\, dw

Convert the radicals to fractional exponets

\int \sqrt[3]{w} + 10\sqrt[5]{w^3}\, dw = \int w^\frac{1}{3} + 10w^\frac{3}{5} \, dw

And then apply inverse power rule

\int \sqrt[3]{w} + 10\sqrt[5]{w^3}\, dw = \frac{3}{4}w^\frac{4}{3} + \frac{25}{4}w^\frac{8}{5} + C

Pauls Online Notes (Calculus - 5.2.9)

Evaluate the following integral

\int \frac{7}{3y^6} + \frac{1}{y^{10}} - \frac{2}{\sqrt[3]{y^4}} \, dy

Convert the radicals to fractional exponents

\int \frac{7}{3y^6} + \frac{1}{y^{10}} - \frac{2}{\sqrt[3]{y^4}} \, dy = \int \frac{7}{3}y^{-6} + y^{-10} - 2y^{\frac{4}{3}} \, dy

And then apply inverse power rule

\int \frac{7}{3y^6} + \frac{1}{y^{10}} - \frac{2}{\sqrt[3]{y^4}} \, dy = -\frac{7}{15}y^{-5} - \frac{1}{9} y^{-9} + 6y^{-\frac{1}{3}} + C

Be careful to make sure you have the right answer.

Common identites

Pauls Online Notes (Calculus - 5.2.14)

Evaluate the following integral

\int \sin x + 10 \csc^2 x \, dx

Simply use linearity to break up the integral, and then apply identities

\int \sin x + 10 \csc^2 x \, dx = -\cos x - 10 \cot x + C

Pauls Online Notes (Calculus - 5.2.15)

Evaluate the following integral

\int 2\cos x - \sec x\tan x \, dx

Simply use linearity to break up the integral, and then apply identities

\int 2\cos x - \sec x\tan x \, dx = 2 \sin x - \sec x + C

Pauls Online Notes (Calculus - 5.2.16)

Evaluate the following integral

\int 12 + \csc \theta \left( {\sin \theta + \csc \theta } \right)\,d\theta

Multiply out and simplify the equation. \int 12 + \csc \theta \left( {\sin \theta + \csc \theta } \right)\,d\theta = \int 12 + 1 + \csc^2 \theta \, d\theta

Because

\csc x= \frac{1}{\sin x}

So we get the answer

\int 12 + \csc \theta \left( {\sin \theta + \csc \theta } \right)\,d\theta = 13 \theta + \cot \theta + C

Pauls Online Notes (Calculus - 5.2.20)

Evaluate the following integral

\int{{\frac{1}{{1 + {x^2}}} + \frac{{12}}{{\sqrt {1 - {x^2}} }}\,dx}}

Using the inverse trig identities

\int{{\frac{1}{{1 + {x^2}}} + \frac{{12}}{{\sqrt {1 - {x^2}} }}\,dx}} = \arctan x + 12 \arcsin x + C