This section introduces the concept of antidifferentiation and linearity.
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.
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.
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
Examples
Some example problems from various websites
Inverse power rule
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
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
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
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