# 10. Reduction Formulae

by M. Bourne

You may have noticed in the Table of Integrals that some integrals are given in terms of a simpler integral. These require a few steps to find the final answer.

**Reduction formulae** are integrals involving some variable `n`, as well as the usual `x`. They are normally obtained from using integration by parts.

We use the notation `I_n` when writing reduction formulae.

### Example 1

Given the reduction formula

`I_n=intsin^nx\ dx` `=-1/ncos x\ sin^(n-1)x+(n-1)/nI_(n-2),`

find `intsin^4x\ dx`.

Answer

Applying the reduction formula with `n = 4` gives:

`int sin^4x\ dx` `=-1/4cos x\ sin^3x+3/4I_2`

We now need to find `I_2=intsin^2x\ dx`, which corresponds to `n=2`.

Now, from our table of integrals,

`intsin^2xdx` `=x/2-1/2sin x\ cos x+K`

So putting it all together gives:

`intsin^4x\ dx`

`=-1/4cos x\ sin^3x` `+3/4[x/2-1/2sin x\ cos x+K]`

`=-1/4cos x\ sin^3x` `+3/8x` ` -3/8 cos x\ sin x+K^'`

`=3/8 x - 1/4 sin(2 x)` `+ 1/32 sin(4 x)+K^'` (afer some manipulation)

**NOTE: **We used `K` and `K^'` since the value of those constants is actually different.

For interest, here are the graphs of the function `y=sin^4 x` and its integral.

Graph of `y(x)=sin^4 x`, and its integral `y=3/8 x - 1/4 sin(2 x)+ 1/32 sin(4 x)` (I've used `K=0`).

### Example 2

We are given that if

`I_n=inttan^nx\ dx`,

then `I_n=1/(n-1)tan^(n-1)x-I_(n-2).`

Find `inttan^3x\ dx`.

Answer

`inttan^3x\ dx=1/2tan^2x-I_1`

Now `I_1=inttan x\ dx` `=-ln\ |cos x|+K`

So `inttan^3x\ dx` `=1/2tan^2x+ln\ |cos x|+K`

**Note: **Often in this work there is more than one way to attack the question, so it is possible to get a correct answer, but in a different form. Using math software, like Scientific Notebook, you may get:

`inttan^3x\ dx` `=1/2tan^2x-1/2ln\ |1+tan^2x|`

Are the 2 answers the same?

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