Techniques of Integration

We have already learnt how to differentiate a function and how to take the derivative from the previous article. Is there a way for us to get back the original equation? Yes, there is. In this article, we will show you how to get back the original equation from its derivative using integration – a topic that studies the relationship between a whole and its parts.

Integration is a fundamental topic that can be found in many mathematics syllabi such as the IGCSE Additional Mathematics (0606) and Singapore SEAB Additional Mathematics.

Basics of Integration

Many students find the idea of integration to be a bit intimidating. Here’s a slightly more sophisticated way to think about integration: as the inverse of differentiation. It’s a useful trick, and it’s easy enough to remember. Integration is an operation we use all the time, to find areas under graphs, the volume of a solid, and many others.

Integration of Power Functions

The power rule is the only integration rule that you’ll use when you’re first learning how to integrate. The power rule for integration is a way to find the integral of powers of $x$. The power rule for integration is applicable only to monomials, polynomials without any brackets.

$\int{{{x}^{n}}\text{d}x=\frac{{{x}^{n+1}}}{n+1}}+c,\text{ }n\ne -1$

$\int{{{(ax+b)}^{n}}\text{d}x=\frac{{{(ax+b)}^{n+1}}}{a(n+1)}}+c$

The rule is simple.

1. Insert the expression that you’re integrating, ($x$ or $ax+b$) and add one to the exponent: $n+1$
2. Divide the whole thing by that same value $n+1$
Note: For $ax+b$, we have to remember to divide with the coefficient of $x$ too.
3. Finally, don’t forget to add the constant $c$.

It’s not a difficult concept to grasp, but sometimes it’s a little confusing. If you’re having trouble understanding it, let’s try some practice questions together.

Practice Question 1

Integrate the following expressions with respect to $x$:

(a)

$10{{x}^{4}}-5{{x}^{2}}+x+3$

(a) $10{{x}^{4}}-5{{x}^{2}}+x+3$

(b)

$6\sqrt[3]{x}-\frac{2}{\sqrt{x}}+8{{x}^{3}}$

(b) $6\sqrt[3]{x}-\frac{2}{\sqrt{x}}+8{{x}^{3}}$

(c)

$\frac{4\sqrt[5]{x}+3{{x}^{4}}+\sqrt{x}}{6{{x}^{3}}}$

(c) $\frac{4\sqrt[5]{x}+3{{x}^{4}}+\sqrt{x}}{6{{x}^{3}}}$

\begin{aligned} & \int{10{{x}^{4}}-5{{x}^{2}}+{{x}^{1}}+3\,\text{d}x} \\ & =\frac{10{{x}^{5}}}{5}-\frac{5{{x}^{3}}}{3}+\frac{{{x}^{2}}}{2}+3x+c \\ & =2{{x}^{5}}-\frac{5}{3}{{x}^{3}}+\frac{1}{2}{{x}^{2}}+3x+c \\ \end{aligned}

\begin{aligned} & \int{6{{x}^{\frac{1}{3}}}-\frac{2}{{{x}^{\frac{1}{2}}}}+8{{x}^{3}}}\,\text{d}x \\ & =\int{6{{x}^{\frac{1}{3}}}-2{{x}^{-\frac{1}{2}}}+8{{x}^{3}}}\,\text{d}x \\ & =\frac{6{{x}^{\frac{4}{3}}}}{\frac{4}{3}}-\frac{2{{x}^{\frac{1}{2}}}}{\frac{1}{2}}+\frac{8{{x}^{4}}}{4}+c \\ & =\frac{9}{2}{{x}^{\frac{4}{3}}}-4{{x}^{\frac{1}{2}}}+2{{x}^{4}}+c \\ & =\frac{9}{2}{{\sqrt[3]{x}}^{4}}-4\sqrt{x}+2{{x}^{4}}+c \\ \end{aligned}

\begin{aligned} & \int{\frac{4\sqrt[5]{x}+3{{x}^{4}}+\sqrt{x}}{6{{x}^{3}}}\text{d}x} \\ & =\int{\frac{4{{x}^{\frac{1}{5}}}}{6{{x}^{3}}}+\frac{3{{x}^{4}}}{6{{x}^{3}}}+\frac{{{x}^{\frac{1}{2}}}}{6{{x}^{3}}}\text{d}x} \\ & =\int{\frac{2}{3}{{x}^{\frac{1}{5}-3}}+\frac{1}{2}}{{x}^{4-3}}+\frac{1}{6}{{x}^{\frac{1}{2}-3}}\text{d}x \\ & =\int{\frac{2}{3}{{x}^{-\frac{14}{5}}}+\frac{1}{2}{{x}^{1}}+\frac{1}{6}{{x}^{-\frac{5}{2}}}\,\text{d}x} \\ & =\frac{2}{3}\left( \frac{{{x}^{-\frac{9}{5}}}}{-\frac{9}{5}} \right)+\frac{1}{2}\left( \frac{{{x}^{2}}}{2} \right)+\frac{1}{6}\left( \frac{{{x}^{-\frac{3}{2}}}}{-\frac{3}{2}} \right)+c \\ & =-\frac{10}{27}{{x}^{-\frac{4}{5}}}+\frac{1}{4}{{x}^{2}}-\frac{1}{9}{{x}^{-\frac{3}{2}}}+c \\ \end{aligned}

Integration of Reciprocal Functions

The reciprocal function is a function whose value is the multiplicative inverse of another value. Integration of reciprocal functions is a powerful extension of the power rule of integration. However, this integral rule does not follow the usual power rule for integrals. So, how do we solve integrals involving reciprocal functions?

The following formula is used when a reciprocal function is involved in integration.

$\int{\frac{1}{x}\text{ d}x=\ln x}+c$

$\int{\frac{1}{ax+b}\text{ d}x=\frac{1}{a}\ln (ax+b)}+c$

We will obtain a natural logarithm as the answer, instead of a monomial.

Practice Question 2

Integrate the following expression with respect to $x$.

(a)

$\int{\frac{5}{6x}\text{ d}x}$

(a) $\int{\frac{5}{6x}\text{ d}x}$

(b)

$\int{\frac{1}{4x-3}\text{ d}x}$

(b) $\int{\frac{1}{4x-3}\text{ d}x}$

(c)

$\int{\frac{8+{{\left( 4x-3 \right)}^{5}}}{{{\left( 4x-3 \right)}^{6}}}\text{ d}x}$

(c) $\int{\frac{8+{{\left( 4x-3 \right)}^{5}}}{{{\left( 4x-3 \right)}^{6}}}\text{ d}x}$

(d)

$\int{\frac{5}{2-5x}\text{ d}x}$

(d) $\int{\frac{5}{2-5x}\text{ d}x}$

\begin{aligned} & \int{\frac{5}{6x}\text{d}x} \\ & =\int{\frac{5\cdot 1}{6x}\text{d}x} \\ & =\frac{5}{6}\int{\frac{1}{x}\text{d}x} \\ & =\frac{5}{6}\ln x+c \\ \end{aligned}

\begin{aligned} & \int{\frac{1}{4x-3}\text{d}x} \\ & =\frac{\ln \left( 4x-3 \right)}{4}+c \\ & =\frac{1}{4}\ln \left( 4x-3 \right)+c \\ \end{aligned}

\begin{aligned} & \int{\frac{8}{{{\left( 4x-3 \right)}^{6}}}+\frac{{{\left( 4x-3 \right)}^{5}}}{{{\left( 4x-3 \right)}^{6}}}\text{d}x} \\ & =\int{8{{\left( 4x-3 \right)}^{-6}}+\frac{1}{4x-3}}\,\text{d}x \\ & =\frac{8{{\left( 4x-3 \right)}^{-5}}}{\left( -5 \right)\left( 4 \right)}+\frac{\ln \left( 4x-3 \right)}{4}+c \\ & =-\frac{2}{5}{{\left( 4x-3 \right)}^{-5}}+\frac{1}{4}\ln \left( 4x-3 \right)+c \\ \end{aligned}

\begin{aligned} & \int{\frac{5}{2-5x}\text{d}x} \\ & =5\int{\frac{1}{2-5x}\text{d}x} \\ & =5\frac{\ln \left( 2-5x \right)}{\left( -5 \right)}+c \\ & =-\ln \left( 2-5x \right)+c \\ \end{aligned}

Integration of Exponential Functions

In general, we cannot apply the power rule to the exponent on $e$. But fret not, exponential functions can be integrated using the formulas:

$\int{{{\text{e}}^{x}}\text{ d}x={{\text{e}}^{x}}}+c$

$\int{{{\text{e}}^{ax+b}}\text{ d}x=\frac{1}{a}{{\text{e}}^{ax+b}}}+c$

The integral of this function is easy to calculate as this function may also be solved using integration by parts or the substitution method.

Practice Question 3

Integrate the following expression with respect to $x$.

(a)

$\int{{{\text{e}}^{2x-1}}\text{ d}x}$

(a) $\int{{{\text{e}}^{2x-1}}\text{ d}x}$

(b)

$\int{\frac{{{\text{e}}^{4x}}+5}{{{\text{e}}^{4x}}}\text{ d}x}$

(b) $\int{\frac{{{\text{e}}^{4x}}+5}{{{\text{e}}^{4x}}}\text{ d}x}$

(c)

$\int{\frac{{{\left( {{\text{e}}^{2x}}-{{\text{e}}^{x}} \right)}^{2}}}{{{\text{e}}^{x}}}\text{ d}x}$

(c) $\int{\frac{{{\left( {{\text{e}}^{2x}}-{{\text{e}}^{x}} \right)}^{2}}}{{{\text{e}}^{x}}}\text{ d}x}$

(d)

$\int{\frac{1-{{\text{e}}^{4-3x}}}{{{\text{e}}^{4-3x}}}\text{ d}x}$

(d) $\int{\frac{1-{{\text{e}}^{4-3x}}}{{{\text{e}}^{4-3x}}}\text{ d}x}$

\begin{aligned} & \int{{{\text{e}}^{2x-1}}}\,\text{d}x \\ & =\frac{{{\text{e}}^{2x-1}}}{2}+c \\ \end{aligned}

\begin{aligned} & \int{\frac{{{\text{e}}^{4x}}+5}{{{\text{e}}^{4x}}}}\text{ d}x \\ & =\int{\frac{{{\text{e}}^{4x}}}{{{\text{e}}^{4x}}}+\frac{5}{{{\text{e}}^{4x}}}\,\text{d}x} \\ & =\int{1+5{{\text{e}}^{-4x}}\,\text{d}x} \\ & =x+\frac{5{{\text{e}}^{-4x}}}{\left( -4 \right)}+c \\ & =x-\frac{5}{4}{{\text{e}}^{-4x}}+c \\ \end{aligned}

\begin{aligned} & \int{\frac{{{\left( {{\text{e}}^{2x}}-{{\text{e}}^{x}} \right)}^{2}}}{{{\text{e}}^{x}}}\text{d}x} \\ & =\int{\frac{\left( {{\text{e}}^{2x}}-{{\text{e}}^{x}} \right)\left( {{\text{e}}^{2x}}-{{\text{e}}^{x}} \right)}{{{\text{e}}^{x}}}\text{d}x} \\ & =\int{\frac{{{\left( {{\text{e}}^{2x}} \right)}^{2}}-2{{\text{e}}^{2x}}\cdot {{\text{e}}^{x}}+{{\left( {{\text{e}}^{x}} \right)}^{2}}}{{{\text{e}}^{x}}}}\text{ d}x \\ & =\int{\frac{{{\text{e}}^{4x}}-2{{\text{e}}^{3x}}+{{\text{e}}^{2x}}}{{{\text{e}}^{x}}}\text{d}x} \\ & =\int{\frac{{{\text{e}}^{4x}}}{{{\text{e}}^{x}}}-\frac{2{{\text{e}}^{2x}}}{{{\text{e}}^{x}}}+\frac{{{\text{e}}^{2x}}}{{{\text{e}}^{x}}}\text{d}x} \\ & =\int{{{\text{e}}^{4x-x}}-2{{\text{e}}^{3x-x}}+{{\text{e}}^{2x-x}}\text{d}x} \\ & =\int{{{\text{e}}^{3x}}-2{{\text{e}}^{2x}}+{{\text{e}}^{x}}\,\text{d}x} \\ & =\frac{{{\text{e}}^{3x}}}{3}-\frac{2{{\text{e}}^{2x}}}{2}+{{\text{e}}^{x}}+c \\ & =\frac{1}{3}{{\text{e}}^{3x}}-{{\text{e}}^{2x}}+{{\text{e}}^{x}}+c \\ \end{aligned}

\begin{aligned} & \int{\frac{1-{{\text{e}}^{4-2x}}}{{{\text{e}}^{4-3x}}}}\,\text{d}x \\ & =\int{\frac{1}{{{\text{e}}^{4-3x}}}-\frac{{{\text{e}}^{4-3x}}}{{{\text{e}}^{4-3x}}}\text{d}x} \\ & =\int{{{\text{e}}^{-\left( 4-3x \right)}}-1\,\text{d}x} \\ & =\int{{{\text{e}}^{3x-4}}-1\,\text{d}x} \\ & =\frac{{{\text{e}}^{3x-4}}}{3}-x+c \\ \end{aligned}

Integration of Trigonometric Functions

Previously, we have learnt the basics of trigonometric derivatives from the topic Techniques and Applications of Differentiation. Now, let’s look at the integration of trigonometric functions.

$\int{\cos x\,\text{d}x}=\sin x+c$
$\int{\sin x\,\text{d}x}=-\cos x+c$
$\int{{{\sec }^{2}}x\,\text{d}x}=\tan x+c$

$\int{\cos \left( ax+b \right)\text{ d}x=\frac{\sin \left( ax+b \right)}{a}}+c$
$\int{\sin \left( ax+b \right)\text{ d}x=-\frac{\cos \left( ax+b \right)}{a}}+c$
$\int{{{\sec }^{2}}\left( ax+b \right)\text{ d}x=\frac{\tan \left( ax+b \right)}{a}}+c$

Special thing about this is, the integrals and the derivatives are inter-related. Yes, it is the reverse of the derivatives of trigonometric functions.

Practice Question 4

Integrate the following expression with respect to $x$.

(a)

$7\cos 3x+{{\sec }^{2}}x$

(a) $7\cos 3x+{{\sec }^{2}}x$

(b)

$\frac{1}{{{x}^{3}}}-\sin \left( 5x+2 \right)$

(b) $\frac{1}{{{x}^{3}}}-\sin \left( 5x+2 \right)$

(c)

$\sin 3x-{{\sec }^{2}}6x$

(c) $\sin 3x-{{\sec }^{2}}6x$

\begin{aligned} & \int{7\cos 3x+{{\sec }^{2}}x\,\text{d}x} \\ & =\frac{7\sin 3x}{3}+\tan x+c \\ & =\frac{7}{3}\sin 3x+\tan x+c \\ \end{aligned}

\begin{aligned} & \int{\frac{1}{{{x}^{3}}}-\sin \left( 5x+2 \right)\text{d}x} \\ & =\int{{{x}^{-3}}-\sin \left( 5x+2 \right)\text{d}x} \\ & =\frac{{{x}^{-2}}}{-2}-\frac{\left[ -\cos \left( 5x+2 \right) \right]}{5}+c \\ & =-\frac{1}{2}{{x}^{-2}}+\frac{1}{5}\cos \left( 5x+2 \right)+c \\ \end{aligned}

\begin{aligned} & \int{\sin 3x-{{\sec }^{2}}6x}\,\text{d}x \\ & =-\frac{\cos 3x}{3}-\frac{\tan 6x}{6}+c \\ \end{aligned}

The trigonometric identities are special equations used to simplify complex trigonometric expressions. The best thing is that the basic trigonometric functions and their identities can also be used to evaluate integrals involving the trigonometric function. Now, let’s look at the integration of quadratic trigonometric functions.

$\int{{{\cos }^{2}}x\,\text{d}x}=\int{\frac{\cos 2x+1}{2}}\,\text{d}x$
$\int{{{\sin }^{2}}x\text{ d}x}=\int{\frac{1-\cos 2x}{2}\text{ d}x}$
$\int{{{\tan }^{2}}x\text{ d}x}=\int{1-{{\sec }^{2}}x\text{ d}x}$

[Use $\cos 2A=2{{\cos }^{2}}A-1$]
[Use $\cos 2A=1-2{{\sin }^{2}}A$]
[Use ${{\sec }^{2}}A=1+{{\tan }^{2}}A$]

Practice Question 5

Integrate the following expression with respect to $x$.

(a)

$3{{\sec }^{2}}4x+\sin 3x$

(a) $3{{\sec }^{2}}4x+\sin 3x$

(b)

$6\cos 2x-{{\sec }^{2}}2x+\sqrt{x}$

(b) $6\cos 2x-{{\sec }^{2}}2x+\sqrt{x}$

(c)

${{\sin }^{2}}2x+{{\cos }^{2}}x$

(c) ${{\sin }^{2}}2x+{{\cos }^{2}}x$

\begin{aligned} & \int{3{{\sec }^{2}}4x+\sin 3x} \\ & =\frac{3\tan 4x}{4}-\frac{\cos 3x}{3}+c \\ & =\frac{3}{4}\tan 4x-\frac{1}{3}\cos 3x+c \\ \end{aligned}

\begin{aligned} & \int{6{{\cos }^{2}}x-{{\sec }^{2}}}+\sqrt{x}\,\text{d}x \\ & =6\left[ \frac{\sin 2x}{2} \right]-\left[ \frac{\tan 2x}{2} \right]+\frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}}+c \\ & =3\sin 2x-\frac{1}{2}\tan 2x+\frac{2}{3}{{x}^{\frac{3}{2}}}+c \\ \end{aligned}

\begin{aligned} & \int{{{\sin }^{2}}2x+{{\cos }^{2}}x\,\text{d}x} \\ & =\int{\frac{1-{{\cos }^{2}}\left( 2x \right)}{2}}+\frac{\cos 2x+1}{2}\text{d}x \\ & =\int{\frac{1}{2}}-\frac{1}{2}\cos 4x+\frac{1}{2}\cos 2x+\frac{1}{2}\text{d}x \\ & =\int{1-\frac{1}{2}\cos 4x}+\frac{1}{2}\cos 2x\,\text{d}x \\ & =x-\frac{1}{2}\left( \frac{\sin 4x}{4} \right)+\frac{1}{2}\left( \frac{\sin 2x}{2} \right)+c \\ & =x-\frac{1}{8}\sin 4x+\frac{1}{4}\sin 2x+c \\ \end{aligned}

Integration As A Reverse Process of Differentiation

From above, we know that integration and differentiation are inverse operations. Differentiation is a way of finding the rate at which something changes. Integration is a way of finding the total amount of change.

Is it possible to solve an integral question if you are only given a function and its derivative? We can. The answer is surprisingly simple. Let’s have a look at the practice question below.

Practice Question 6

Show that $\frac{\text{d}}{\text{d}x}\left[ (2x-1)\sqrt{x+3} \right]=\frac{6x+11}{2\sqrt{x+3}}$. Hence, find $\int{\,\frac{6x+11}{\sqrt{x+3}}\text{ d}x}$.

\begin{aligned} & \frac{\text{d}}{\text{d}x}\left[ \left( 2x-1 \right){{\left( x+3 \right)}^{\frac{1}{2}}} \right] \\ & =\left( 2x-1 \right)\left[ \frac{1}{2}{{\left( x+3 \right)}^{-\frac{1}{2}}} \right]+{{\left( x+3 \right)}^{\frac{1}{2}}}\left( 2 \right) \\ & =\left( 2x-1 \right)\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{x+3}}+2\sqrt{x+3} \\ & =\frac{2x-1}{2\sqrt{2x+3}}+2\sqrt{x+3} \\ & =\frac{2x-1}{2\sqrt{2x+3}}+\left( 2\sqrt{x+3}\cdot \frac{2\sqrt{x+3}}{2\sqrt{x+3}} \right) \\ & =\frac{2x-1+4\left( x+3 \right)}{2\sqrt{x+3}} \\ & =\frac{2x-1+4x+12}{2\sqrt{x+3}} \\ & =\frac{6x+11}{2\sqrt{x+3}} \\ \end{aligned}

\begin{aligned} \int{\frac{6x+11}{2\sqrt{x+3}}\text{d}x}=&\left( 2x-1 \right)\sqrt{x+3}+{{c}_{1}} \\ \frac{1}{2}\int{\frac{6x+11}{\sqrt{x+3}}\text{d}x}=&\left( 2x-1 \right)\sqrt{x+3}+{{c}_{1}} \\ \int{\frac{6x+11}{\sqrt{x+3}}\text{d}x}=&2\left( 2x-1 \right)\sqrt{x+3}+c;\\&c=2{{c}_{1}} \end{aligned}

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