Integration by Substitution (Edexcel A Level Further Maths: Core Pure): Revision Note

Integrating using trigonometric substitutions

The integrals covered in this revision note are based on the standard results

$\int \frac{1}{\sqrt{a^2-x^2}}\mathrm{d}x=\arcsin \left(\frac{x}{a}\right)+\mathrm{c}$

and

$\int \frac{1}{a^2+x^2}\mathrm{d}x=\frac{1}{a}\mathrm{arc}\tan \left(\frac{x}{a}\right)+\mathrm{c},\mathrm{a}>0,\left|x\right|>\mathrm{a}$

These are given in the formulae booklet

How do I know when to use a trigonometric substitution in integration?

There are three main types of problem

•  Type 1
  Showing the standard results using a substitution ($a$ may have a value)
  The substitution will not be given in such cases
  e.g.  Use a suitable substitution to show that $\int \frac{1}{\sqrt{a^2-x^2}}\mathrm{d}x=\arcsin \left(\frac{x}{a}\right)+c$
  Let $x=a\sin u$, so $\frac{\mathrm{d}x}{\mathrm{d}u}=a\cos u$ and $u=\mathrm{arcsi}n\left(\frac{x}{a}\right)$ $\begin{array}{l}\int \frac{1}{\sqrt{a^2-x^2}}\mathrm{d}x=\int \frac{1}{\sqrt{a^2-{\left(a\sin u\right)}^2}}\left(a\cos u\right)\mathrm{d}u=\int \frac{a\cos u}{\sqrt{a^2(1-{\sin }^{2}u)}}\mathrm{d}u=\int 1\mathrm{d}u=u+c\\ \therefore \int \frac{1}{\sqrt{a^2-x^2}}\mathrm{d}x=\arcsin \left(\frac{x}{a}\right)+c\end{array}$
  The general idea in these types of problems is to reduce the denominator to a single term, often involving the identity , so it can be integrated using standard results or techniques

• Type 2 Reverse chain rule, possibly involving some factorising in the denominator and using ‘adjust’ and ‘compensate’ if necessary
  e.g.  Find $\int \frac{3}{9^2+{\left(2x\right)}^2}\mathrm{d}x$ $\int \frac{3}{9^2+{\left(2x\right)}^2}\mathrm{d}x=3\times \frac{1}{2}\int \frac{2}{9^2+{\left(2x\right)}^2}\mathrm{d}x=\frac{3}{2}\left(\frac{1}{9}\arctan \right(\frac{2x}{9}\left)\right)+c=\frac{1}{6}\arctan \left(\frac{2x}{9}\right)+c$

• Type 3 The denominator contains a three-term quadratic expression – i.e. there is an x term
  In such cases complete the square and use reverse chain rule
  e.g.  Find $\int \frac{1}{25+x^2-6x}\mathrm{d}x$ $\int \frac{1}{25+x^2-6x}\mathrm{d}x=\int \frac{1}{25+{(x-3)}^2-9}\mathrm{d}x=\int \frac{1}{4^2+{(x-3)}^2}\mathrm{d}x=\frac{1}{4}\arctan \left(\frac{x-3}{4}\right)+c$
  (This works since $\frac{\mathrm{d}}{\mathrm{d}x}[x-3]=1$, so effectively there is no reverse chain rule involved) 

• A fourth type of problem may involve a given substitution but the skills to solve these are covered in the A Level Mathematics course

How do I use a trigonometric substitution to find integrals?

• STEP 1
  Identify the type of problem and if a substitution is required
  Determine the substitution if needed

• STEP 2
  For Type 1 problems, differentiate and rearrange the substitution; change everything in the integral
  For Type 2 problems, ‘adjust’ and ‘compensate’ as necessary
  For Type 3 problems complete the square

• STEP 3
  Integrate using standard techniques and results, possibly from the formulae booklet
  For definite integration, a calculator may be used but look out for exact values being required, a calculator may give an approximation

• STEP 4
  Substitute the original variable back in if necessary – this shouldn’t be necessary for definite integration
  For indefinite integration, simplify where obvious and/or rearrange into a required format

Why is arccos x not involved in any of the integration results?

• $\frac{\mathrm{d}}{\mathrm{d}x}\left[\arccos \right(\frac{x}{a}\left)\right]=-\frac{1}{\sqrt{a^2-x^2}}$
  For integration the "-" at the start can be treated as the constant "-1" and so integrating would lead to "-arcsin ..."

  • i.e. $\int -\frac{1}{\sqrt{a^2-x^2}}\mathrm{d}x=-\arcsin \left(\frac{x}{a}\right)+c$

Integrating using hyperbolic substitutions

The integrals covered in this revision note are based on the standard results

$\int \frac{1}{\sqrt{a^2+x^2}}\mathrm{d}x=\mathrm{arsinh}\left(\frac{x}{a}\right)+c$

and

$\int \frac{1}{\sqrt{x^2-a^2}}\mathrm{d}x=\mathrm{arcosh}\left(\frac{x}{a}\right)+c,x>a$ 

These are given in the formulae booklet

How do I know when to use a hyperbolic substitution in integration?

There are three main types of problem

• Type 1
  Showing the standard results using a substitution ($a$ may have a value)
  The substitution will not be given in such cases
  e.g.  Use a suitable substitution to show that $\int \frac{1}{\sqrt{x^2-a^2}}\mathrm{d}x=\mathrm{ar}\cos \text{h}\left(\frac{x}{a}\right)+c$
  Let $x=a\cosh u$, so $\frac{\mathrm{d}x}{\mathrm{d}u}=a\sinh u$ and $u=\mathrm{arcosh}\left(\frac{x}{a}\right)$ $\int \frac{1}{\sqrt{x^2-a^2}}\mathrm{d}x=\int \frac{1}{\sqrt{{\left(a\cosh u\right)}^2-a^2}}\left(a\sinh u\right)\mathrm{d}u=\int \frac{a\sinh u}{\sqrt{a^2({\cosh }^{2}u-1)}}\mathrm{d}u=\int 1\mathrm{d}u=u+c$ $\therefore \int \frac{1}{\sqrt{x^2-a^2}}\mathrm{d}x=\mathrm{arcosh}\left(\frac{x}{a}\right)+c$ The general idea in these types of problems is to reduce the denominator to a single term, often involving the identity ${\cosh }^{2}x-{\sinh }^{2}x\equiv 1$, so it can be integrated using standard results or techniques

• Type 2
  Reverse chain rule, possibly involving some factorising in the denominator and using ‘adjust’ and ‘compensate’ if necessary
  e.g.  Find $\int \frac{3}{\sqrt{4x^2+9}}\mathrm{d}x$ $\int \frac{3}{\sqrt{{\left(2x\right)}^2+3^2}}\mathrm{d}x=3\times \frac{1}{2}\int \frac{2}{\sqrt{{\left(2x\right)}^2+3^2}}\mathrm{d}x=\frac{3}{2}\left(\mathrm{arsinh}\right(\frac{2x}{3}\left)\right)+c=\frac{3}{2}\mathrm{arsinh}\left(\frac{2x}{3}\right)+c$

• Type 3
  The denominator contains a three-term quadratic expression – i.e. there is an $x$ term
  In such cases complete the square and use reverse chain rule
  e.g.  Find $\int \frac{1}{\sqrt{x^2-6x+25}}\mathrm{d}x$ $\int \frac{1}{\sqrt{x^2-6x+25}}\mathrm{d}x=\int \frac{1}{\sqrt{{(x-3)}^2-9+25}}\mathrm{d}x=\int \frac{1}{\sqrt{{(x-3)}^2+4^2}}\mathrm{d}x=\mathrm{arsinh}\left(\frac{x-3}{4}\right)+c$ (This works since $\frac{\mathrm{d}}{\mathrm{d}x}[x-3]=1$,so effectively there is no reverse chain rule involved

• A fourth type of problem may involve a given substitution but the skills to solve these are covered in the A Level Mathematics course, although hyperbolic functions are not

How do I use a hyperbolic substitution to find integrals?

• STEP 1
  Identify the type of problem and if a substitution is required
  Determine the substitution if needed

• STEP 2
  For Type 1 problems, differentiate and rearrange the substitution; change everything in the integral
  For Type 2 problems, ‘adjust’ and ‘compensate’ as necessary
  For Type 3 problems complete the square

• STEP 3
  Integrate using standard techniques and results, possibly from the formulae booklet
  For definite integration, a calculator may be used but look out for exact values being required, a calculator may give an approximation

• STEP 4
  Substitute the original variable back in if necessary – this shouldn’t be necessary for definite integration
  For indefinite integration, simplify where obvious and/or rearrange into a required format

Is artanh x involved in integration?

• The standard result, given in the formulae booklet is
  $\int \frac{1}{a^2-x^2}\mathrm{d}x=\frac{1}{a}\mathrm{artanh}\left(\frac{x}{a}\right)+c,\left|x\right|<a$
  with the alternative result $\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+c$ also given

• Problems involving these often involve partial fractions (since $a^2-x^2$ is the difference of two squares) leading to the 'ln' result

• If you happen to recognise the integral and can use the formulae booklet result involving "artanh" to solve a problem, then do so!