Today’s resource contains proofs of standard integrals. You can download a complete sheet of the standard rules if you are a site member.
In mathematics, an integral is a way to describe a continuous quantity that can be represented as the area under a curve. It is a good idea to become familiar with the basic integrals as these can be directly quoted, especially when tackling harder questions.
Basic Integration Rules
The first rule that you will probably learn is:
$$
\int x^n d x=\frac{x^{n+1}}{n+1}+c \quad \text { where } n \neq-1
$$
In the case above, the value of n cannot be equal to -1. If n = -1 then we would be dividing by zero in the result.
We can extend the above result and consider the following:
$$
\begin{aligned}
& \int a x^n \space d x=\frac{a x^{n+1}}{n+1}+c \quad \text { where } n \neq-1 \
\end{aligned}
$$
Another typical rule that is taught fairly early is:
$$
\begin{aligned}
& \int \frac{1}{x} \space d x=\ln\left|x\right|+c
\end{aligned}
$$
You may see this written as:
$$
\begin{aligned}
& \int \frac{dx}{x}=\ln\left|x\right|+c
\end{aligned}
$$
Trigonometric Integrals
Before learning integration, you should have been taught differentiation. The following rules should, therefore, be self-explanatory. Some of the first rules you will learn are:
$$
\begin{aligned}
& \int \sin x \space d x=-\cos x+c \
\end{aligned}
$$
and
$$
\begin{aligned}
& \int \cos x \space d x=\sin x+c
\end{aligned}
$$
When first learning integration, you may be shown that coefficients can be taken outside the integral as follows:
$$
\begin{aligned}
\int 3 \sin x \space d x & =3 \int \sin x \space d x \\\\
& =3(-\cos x)+c \\\\
& =-3 \cos x+c
\end{aligned}
$$
Similarly:
$$
\begin{aligned}
\int -5 \sin x \space d x & =-5 \int \sin x \space d x \\\\
& =-5(-\cos x)+c \\\\
& =5 \cos x+c
\end{aligned}
$$
From your knowledge of differentiation, you should also be aware of the following rules:
$$
\begin{aligned}
\int \sec^2 x \space d x &=\tan x+c\\\\\\
\int \csc^2 x \space d x &=-\cot x+c\\\\\\
\int \sec x\tan x \space d x &=\sec x + c\\\\\\
\int \csc x\cot x \space d x &=-\csc x + c\\\\\\
\end{aligned}
$$
Integration of \(\tan x\)
The next integral is derived through substitution:
$$
\begin{aligned}
\int \tan x \space d x & =\int \frac{\sin x}{\cos x} \space d x \\\\
\text { Let } u & =\cos x \\\\
\frac{d u}{d x} & =-\sin x \\\\
d u & =-\sin x \space d x
\end{aligned}
$$
Now substitute
$$\begin{aligned}
\int \frac{\sin x}{\cos x} \space d x & =-\int \frac{-\sin x}{\cos x} \space d x \\\\
& =-\int \frac{d u}{u} \\\\
& =-\ln |u|+c \\\\
& =-\ln |\cos x|+c
\end{aligned}
$$
Note that there is an alternative form for this result:
$$
\begin{aligned}
\int \tan x \space d x & =-\ln |\cos x|+c\\\\
&=\ln |(\cos x)^{-1}|+c \quad \quad \text {by applying law of logs}\\\\
&=\ln \left|\frac{1}{\cos x}\right|+c\\\\
&=\ln |\sec x|+c
\end{aligned}
$$
Integration of \(\cot x\)
You can use a similar method to the one above:
$$
\begin{aligned}
\int \cot x \space d x & =\int \frac{\cos x}{\sin x} \space d x \\\\
\text { Let } u & =\sin x \\\\
\frac{d u}{d x} & =\cos x \\\\
d u & =\cos x \space d x
\end{aligned}
$$
Now substitute
$$\begin{aligned}
\int \frac{\cos x}{\sin x} \space d x & =\int \frac{du}{u} \space d x \\\\
& =\ln |u|+c \\\\
& =\ln |\sin x|+c
\end{aligned}
$$
Integration of \(\sec x\)
The first step involves multiplying by \(\sec x + \tan x \). Since this is done in the numerator and denominator, the result is equivalent to multiplying by 1 i.e. we are not changing the integral’s value.
$$
\begin{aligned}
\int \sec x \space d x&=\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \space d x \\\\
&=\int \frac{\sec ^2 x+\sec x \tan x}{\sec x+\tan x} \space d x \\\\
\text { Let } u&=\sec x+\tan x \\\\
\frac{d u}{d x}&=\sec x \tan x+\sec ^2 x \\\\
d u&=\left(\sec x \tan x+\sec ^2 x\right) d x \\\\
\int \frac{\sec ^2 x+\sec x \tan x}{\sec x+\tan x} \space d x&=\int \frac{d u}{u} \\\\
&= \ln |u|+c \\\\
&=\ln |\sec x+\tan x|+C
\end{aligned}
$$
Integration of \(\csc x\)
We can use a similar method as above. You really need to know your rules for differentiation in order to follow the steps.
$$
\begin{aligned}
\int \csc x \space d x&=\int \frac{\csc x(\csc x+\cot x)}{\csc x+\cot x} \space d x \\\\
& =\int \frac{\csc ^2 x+\csc x \cot x}{\csc x+\cot x} \space d x \\\\
\text { Let } u&=\csc x+\cot x \\\\
\frac{d u}{d x}&=-\csc x \cot x-\csc ^2 x \\\\
d u&=-\left(\csc x \cot x+\csc ^2 x\right) d x \\\\
\int \frac{\csc ^2 x+\csc x \cot x}{\csc x+\cot x} \space d x&=-\int \frac{d u}{u} \\\\
& =-\ln |u|+c \\\\
& =-\ln |\csc x+\cot x|+c
\end{aligned}
$$
Integration Rules Questions
If you need to practise some basic integration questions then try the integration worksheets with 100 questions and answers.