# Maths Extension 2:Integration

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## Integration by Substitution

Examples: For, $\int\sqrt{a^2 + x^2}\,dx$ use the substitution x = tanθ.

## Partial Fractions & Rational Functions

### Handy Partial Fractions Trick

E.g. Convert $\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\,$ into partial fraction form.

You start off with letting $\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)} \equiv \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\,$

One way to solve the problem is to multiply both sides to get two polynomials and equate the coefficients and solve a few simultaneous equations. The other way which is not taught in HSC but is useful to know is the Residue Theorem. This is somewhat related to Remainder/Factor theorem.

To use this, you simply multiply a factor in the denominator and then substitue the "zero" into the fraction, so:

• To find $A\,$, substitue $x=1\,$ in $(x-1)\times\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\,$ so you get
• $A = \frac{15(1)^2-56(1)+47}{((1)-2)((1)-3)}=\frac{6}{2}=3\,$
• To find $B\,$, substitue $x=2\,$ in $(x-2)\times\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\,$ so you get
• $B = \frac{15(2)^2-56(2)+47}{((2)-1)((2)-3)}=\frac{-5}{-1}=5\,$
• To find $C\,$, substitue $x=3\,$ in $(x-3)\times\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)}\,$ so you get
• $C = \frac{15(3)^2-56(3)+47}{((3)-1)((3)-2)}=\frac{14}{2}=7\,$

Hence $\frac{15x^2-56x+47}{(x-1)(x-2)(x-3)} \equiv \frac{3}{x-1}+\frac{5}{x-2}+\frac{7}{x-3} \,$

This method only works if the factor you are working with looks like $x-a\,$ and not $x^n+a\,$ or $(x-a)^n\,$.

## Integration by Parts (Reverse Product Rule)

Integration by parts allows the student to perform integration of the form $\int f(x)g(x)dx$ with relative ease.

Students of Mathematics Extension 2 would already be familiar with the rule for differentiating the product of two functions: $\frac{d}{dx}f(x)g(x) = f'(x)g(x) + f(x)g'(x)$. Integration by parts uses this fact to construct a rule applicable to integration, as follows:

Suppose G'(x) = g(x); that is, suppose the primitive of g is G. Thus, by the product rule and recasting:

$\frac{d}{dx}f(x)G(x)= f'(x)G(x) + f(x)g(x)$

$f(x)g(x) = \frac{d}{dx}f(x)G(x) - f'(x)G(x)$

Integrating both sides gives:

$\int f(x)g(x)dx$  $=\int \frac{d}{dx} f(x)G(x) - \int f(x)G(x)dx$
$=f(x)G(x) - \int f'(x)G(x)dx$

which is the rule of integration by parts.

## Reduction Formulae

For example:

Find $\int sin^nx\, dx$

and hence find $\int_{0}^{\frac{\Pi}{4}} sin^9x\, dx$

Solution:

Let
$\int sin^nx=I_n \,$

And let
$sin^nx = sin^{n-1}*sinx\,$

Integrating by parts
$u = sin^nx \,$
$u'= (n-1)(sin^{n-2}x)(cosx) \,$

$v'=sinx\,$
$v = -cosx\,$

NOTE: To evaluate $u'\,$, use the chain rule.

$I_n = uv-\int u'v\, dx$

$I_n = -sin^nxcosx-\int (n-1)(sin^{n-2}x)(cosx)(-cosx)\, dx$

Take a factor of (n − 2) out of the integral.

$I_n = -sin^nxcosx+(n-1)\int (sin^{n-2}x)(cos^2x)\, dx$

Using $sin^2x+cos^2x=1\,$ substitute $cos^2x=(1-sin^2x)\,$

$I_n = -sin^nxcosx+(n-1)\int (sin^{n-2}x)(1-sin^2x)\, dx$

$\int (sin^{n-2}x)(1-sin^2x)\, dx$ $= \int (sin^{n-2}x) - \int (sin^nx)\,$ (by expanding and separating)

$I_n = -sin^nxcosx+(n-1)\int (sin^{n-2}x) - (n-1)\int (sin^nx)\,$

Add $(n-1)\int sin^nx\,$ to both sides.

For the Left side you have:

$\int (sin^nx)+n\int (sin^nx)-\int (sin^nx)\,$
$= n\int (sin^nx)\,$

$n\int (sin^nx) = -sin^nxcosx+(n-1)\int (sin^{n-2}x)\,dx$

Divide by n

$\int (sin^nx) = \frac{-sin^nxcosx}{n}+\frac{n-1}{n}\int (sin^{n-2}x)\,dx$

This is our general solution for $\int sin^nx \,$
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