# Part 9: Harder Standard Integrals | Beginner’s Guide to Year 12 Ext 2 Maths

Need confidence approaching harder standard integrals? In this article, we discuss strategies and provide practice questions and solutions for you to hone your skills.

Integration is a complex process. Extension 2 Maths presents you with harder standard integrals to solve. In this article, we discuss the sorts of questions you will face, how to tackle them, and provide questions to test yourself with.

## Practise a wide range of Harder Standard Integrals questions

This worksheet has 4 levels of difficulty to test your knowledge

## Year 12 Mathematics Extension 2: Harder Standard Integrals

Understanding and mastering integration techniques play a crucial role for later 4-unit topics as integration often comes up in various topics.

Integration is not a simple process, while we can differentiate any function, we cannot simply integrate any function.

Hence, it becomes essential for students to recognise the type of given integral before any steps are taken to solving it.

### NESA Syllabus Outcomes

NESA has provided the following outcomes for this topic:

• Understand and uses different representations of numbers and functions to model, prove results and find solutions to problems in a variety of contexts
• Applies techniques of integration to structured and unstructured problems
• Applies various mathematical techniques and concepts to model and solve structured, unstructured and multi-step problems
• Communicates and justifies abstract ideas and relationships using appropriate language, notation and logical argument

## Assumed Knowledge

Students should have a strong understanding of integration and the various types of methods used for solving them from 3 unit.

This includes being familiar with all standard integrals from the reference sheet and being able to recognise them when presented in an integration question.

Readers are recommended to revise basic mathematical concepts such as rationalising fractions, splitting improper fractions and completing the square as we will be using them throughout this article.

## Harder Standard Integrals

Standard integrals have been covered extensively in the Maths Advanced and Ext 1 course.

In Ext 2 we will be covering some harder standard integrals as these can sometimes be tricky or difficult to recognise as standard integrals.

Special techniques are often required to transform or simplify them into a standard integral form we already know before evaluating them.

Throughout this article, you will often find many of these techniques are counter-intuitive and usually makes the integral messy, but after some simplification it will eventually transform into a standard integral.

Let us begin with one of the most counter-intuitive techniques:

## Splitting Fractions

### Example 1:

Evaluate

$$\int \frac{1}{1+e^x}dx$$

### Solution:

At first glance, it becomes obvious that this is not a standard integral.

However, by simply adding and subtracting $$e^x$$ to the numerator and splitting the fraction, we obtain our answer.

\begin{align*}
\int \frac{1}{1+e^x} \ dx &= \int \frac{1+e^x-e^x}{1+e^x}dx\\
&= \int \left( 1- \frac{e^x}{1+e^x} \right)dx\\
&= x-ln(1+e^x)+c\\
\end{align*}

## Completing the Square

### Example 1:

Evaluate $$\int \frac{dx}{\sqrt{4x-x^2}}$$

### Solution:

First, by completing the square inside the square root and integrating directly we get

\begin{align*}
\int \frac{dx}{\sqrt{4x-x^2}} &= \int \frac{dx}{\sqrt{4-(x-2)^2}}\\
&= sin^{-1} \frac{x-2}{2} +c\\
\end{align*}

### Example 2:

Evaluate $$\int \frac{dx}{9x^2-6x+2}$$

### Solution:

Again, completing the square and then integrating directly:

\begin{align*}
\int \frac{dx}{9x^2-6x+2} &= \int \frac{dx}{(3x-1)^2+1}\\
&= \frac{1}{3}tan^{-1}(3x-1)+c\\
\end{align*}

## Substitution and Trigonometric Substitution

### Example 1:

Evaluate $$\int \frac{x}{\sqrt{1+x^2}}dx$$

### Solution:

Noting the fact that the derivative of $$1+x^2$$ is $$2x$$, we can let $$u=1+x^2$$ so that $$du=2x \ dx$$.

\begin{align*}
\int \frac{x}{\sqrt{1+x^2}}dx &= \frac{1}{2} \int \frac{1}{\sqrt{u}} \ du\\
&= \sqrt{u} + c\\
&= \sqrt{1+x^2} + c\\
\end{align*}

### Example 2:

Evaluate $$\int \frac{x}{4x^2+9} \ dx$$

### Solution:

Noting the fact that the derivative of $$4x^2+9$$ is $$8x$$, we can let $$u=4x^2+9$$ so that $$du=8x \ dx$$.

\begin{align*}
\int \frac{x}{4x^2+9} \ dx &= \frac{1}{8} \int \frac{1}{u} \ du\\
&= \frac{1}{8} ln(u) +c\\
&= \frac{1}{8}ln(4x^2+9)+c\\
\end{align*}

Note to Students: In Ext 2, it is recommended for students who can recognise the integrals in Example 4 and 5 as standard integrals, to skip the substitution step and go directly to the answer.

This will save time during the exam and students can always come back to check their answers with the substitution method.

### Example 3:

Evaluate $$\int sin^2x \ cosx \ dx$$

### Solution:

Noting the fact that the derivative of $$sinx$$ is $$cosx$$. Let $$u=sinx$$ then $$du=cosx \ dx$$.

\begin{align*}
\int sin^2x \ cosx \ dx &= \int u^2 \ du\\
&= \frac{1}{3}u^3 +c\\
&= \frac{1}{3}sin^3x +c\\
\end{align*}

## Rationalising the Numerator

### Example 1:

Evaluate $$\int \sqrt{ \frac{1-x}{1+x} } \ dx$$

### Solution:

We have always been taught to rationalise the denominator so that we can simplify the fraction. But now we want to rationalise the numerator? Sometimes we will see that by keeping the surd in the denominator, it will slowly turn into a standard integral.

 \begin{align*} \int \sqrt{ \frac{1-x}{1+x} } &= \int \sqrt{ \frac{1-x}{1+x} \times \frac{1-x}{1-x} } \ dx\\ &= \int \frac{1-x}{ \sqrt{1-x^2}} \ dx\\ &= \int \left( \frac{1}{\sqrt{1-x^2}} – \frac{x}{\sqrt{1-x^2}} \right) \ dx\\ &= sin^{-1}x + \sqrt{1-x^2} +c\\ \end{align*}

### Example 2: Harder Splitting Fractions

Evaluate $$\int \frac{x^2+2x+2}{x^2+4x+5} \ dx$$

### Solution:

For this integral, we will be combining many different techniques from previous examples to transform it into a standard integral. Our primary goal here is to keep adding and subtracting terms from the numerator to transform the integral to one we already know how to solve.

 \begin{align*} \int \frac{x^2+2x+2}{x^2+4x+5} \ dx &= \int \frac{x^2+4x+5 – 2x – 3}{x^2+4x+5} \ dx\\ &= \int \left( 1 – \frac{2x+4-1}{x^2+4x+5} \right) \ dx\\ &= \int dx + \int \frac{2x+4}{x^2+4x+5} \ dx \ – \int \frac{dx}{(x+2)^2 +1} \\ &= x +ln(x^2+4x+5)-tan^{-1}(x+2)+c\\ \end{align*}

## Concept Check Questions

Evaluate the following integrals:

1. $$\int \frac{x-1}{x+1} \ dx$$

2. $$\int \frac{3}{4x^2+4x+2} \ dx$$

3. $$\int \frac{x}{\sqrt{3-x}} \ dx$$

4. $$\int \frac{dx}{x+\sqrt{x}}$$

5. $$\int cos^3x \ sinx \ dx$$

## Concept Check Solutions

1. $$x-2ln|x+1|+c$$

2. $$\frac{3}{2}tan^{-1}(2x+1)+c$$

3. $$– \frac{2}{3} (x+6) \sqrt{3-x} +c$$

4. $$2ln| \sqrt{x} +1| +c$$

5. $$– \frac{1}{4}cos^4 +c$$

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