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.