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# 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.

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.

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