Integration by Reduction Formulae

Suppose you have to ∫ex sin(x)dx.

We use integration by parts to obtain the result, only to come across a small snag:

u = ex; dv/dx = sin x
So, du/dx = ex; v = -cos x
∫ex sin(x)dx = -excos x + ∫ excos x dx 1
Now, we have to repeat the integration process for ∫ excos x dx, which is as follows:
u = ex; dv/dx = cos x
du/dx = ex; v = sin x
So, ∫ excos x dx = exsin x - ∫ exsin x dx 2
As you can see, now we are heading towards a never-ending loop; therefore, we have to take a different approach.
Let's make ∫ excos x dx the subject from both equations and equate them.
From 1 => ∫ex sin(x)dx + excos x = ∫ excos x dx 3
From 2 => ∫ excos x dx = exsin x - ∫ exsin x dx 4
Since 3 = 4 => ∫ex sin(x)dx + excos x = exsin x - ∫ exsin x dx
2∫ex sin(x)dx = exsin x - excos x
∫ex sin(x)dx = 1/2ex[sin x - cos x]
We manage to find the integral, yet the sign of integration still remains a part of the solution!

There are plenty of instances where the process of integration is neither straight forward nor easy to come up with. Integration by Reduction Formulae is one such method.

Integration by Reduction Formulae

In this method, we gradually reduce the power of a function up until it comes to a stage that it can be integrated. This is usually accomplished by integration by parts method.

E.g.

∫ [ln x]ndx

Let's use the integration by parts method:
u = [ln x]n => du/dx = n/x[ln x]n-1; dv/dx = 1 => v = x
∫ [ln x]n dx = x[ln x]n - n/x∫x[ln x]n-1
= x[ln x]n - n∫[ln x]n-1
If ∫ [ln x]ndx = In, then ∫[ln x]n-1 = In-1
Therefore,
In = x [ln x]n - n In-1

This is the reduction formula for integrating [ln x]n with respect to x. It doesn't produce any result at this stage; therefore. let's see how it really works.

Suppose you want to find ∫[ln x]3 dx, which is I3.
I3 = x[ln x]3 - 3 I2]
= x(ln x)3 - 3 [x (ln x)2 - 2 I1]
= x(ln x)3 - 3x (ln x)2 + 6 I1
= x(ln x)3 - 3x (ln x)2 + 6 [x (ln x)1 - I0]
= x(ln x)3 - 3x (ln x)2 + 6x (ln x)1 - 6I0
= x(ln x)3 - 3x (ln x)2 + 6x (ln x) - 6[x(ln x)0 - 0 I-1]
= x(ln x)3 - 3x (ln x)2 + 6 x (ln x) - 6x + c
So, ∫[ln x]3 dx = x(ln x)3 - 3x (ln x)2 + 6x (ln x) - 6x + c

Now, let's prove that the integral obtained by the reduction formula is, indeed, that of ∫[ln x]3 dx, by differentiating the former:

Proof

Let' apply the product rule and the chain rule in differentiating it
d[x(ln x)3 - 3x (ln x)2 + 6x (ln x) - 6x]/dx = x(3)[ln x]2(1/x) + [ln x]3 - 6ln x - 3 [ln x]2 + 6 + 6 ln x -6
= [ln x]3

The reduction formulae can be extended to a range of functions. The procedure, however, is not the same for every function.

E.g.1

Find a reduction formula to integrate ∫sinn x dx and hence find ∫sin4 x dx.

∫sinnx dx = ∫sinn-1x sin x dx
Let u = ∫sinn-1x and dv/dx = sin x
So, du/dx = (n-1)sinn-2x cos x; v = -cos x
∫sinnx dx = -sinn-1 x cos x + ∫ cos x (n-1)sinn-2x cos x dx
= -sinn-1x cos x + ∫ cos2 x (n-1)sinn-2x dx
= -sinn-1x cos x + ∫ [1 - sin2x ](n-1)sinn-2x dx
= -sinn-1x cos x + ∫ (n-1)sinn-2x dx - ∫(n-1)sin2x sinn-2x dx
= -sinn-1x cos x + ∫ (n-1)sinn-2x dx - ∫(n-1)sinnx dx
∫n sinnx dx = -sinn-1x cos x + ∫ (n-1)sinn-2x dx
∫sinnx dx = -1/n sinn-1x cos x+ ∫ (n-1)/n sinn-2x dx
In = -1/n sinn-1 x cos x + (n-1)/n In-2

∫ sin4x dx = -1/4 sin3x cos x + ∫ 3/4 sin2x dx
= -1/4 sin3x cos x + 3/4[-1/2 cos x sin x + ∫ 1/2 sin0x dx]
= -1/4 sin3x cos x -3/8 cos x sin x + 3/8 x + c dx
= -1/4 sin3x cos x -3/8 sin x cos x + 3/8 x + c
since sin 2x = 2 sin x cos x
∫ sin4x dx = -1/4 sin3x cos x - 3/16 sin 2x + 3/8 x + c

E.g.2

Find a reduction formula to integrate ∫cosn x dx and hence find ∫cos4 x dx.

∫cosnx dx = ∫cosn-1x cos x dx
Let u = ∫cosn-1x and dv/dx = cos x
So, du/dx = -(n-1)cosn-2x sin x; v = sin x
∫cosnx dx = -cosn-1 x sin x + ∫ sin x (n-1)cosn-2x sin x dx
= cosn-1x sin x + ∫ sin2 x (n-1)cosn-2x dx
= cosn-1x sin x + ∫ [1 - cos2x ](n-1)cosn-2x dx
= cosn-1x sin x + ∫ (n-1)cosn-2x dx - ∫(n-1)cosnx dx
∫n cosnx dx = cosn-1x sin x + ∫(n-1)cosn-2x dx
∫cosnx dx = 1/n cosn-1x sin x + ∫ (n-1)/n cosn-2x dx
In = 1/n cosn-1 x sin x + (n-1)/n In-2

∫ cos4x dx = 1/4 cos3x sin x + ∫ 3/4 cos2x dx
= 1/4 cos3x sin x + 3/4[1/2 cos x sin x + ∫ 1/2 cos0x dx]
= 1/4 cos3x sin x + 3/8 cos x sin x + 3/8 x + c
Since sin 2x = 2 sin x cos x
∫ cos4x dx = 1/4 cos3x sin x + 3/16 sin 2x + 3/8 x + c

E.g.3

Find a reduction formula to integrate ∫ tann x dx and hence find ∫tan4x dx.

∫tann x dx = ∫tann-2x tan2x dx
u = tann-2x => du/dx = n-2 tann-3x sec2x; dv/dx tan2x = sec2x - 1 => v = tan x - x
The integration leads to the following formula:
∫tann x dx = 1/n-1 tann-1 x - ∫tann-2 x dx
In = 1/n-1 In-1 - In-2

∫tan4 x dx = 1/3 tan3 x - ∫tan2 x dx
= 1/3 tan3 x - [tan x - ∫ dx]
= 1/3 tan3 x - tan x + x + c

E.g.4

Find a reduction formula to integrate ∫ exxn x dx and hence find ∫exx3dx.

u = xn => du/dx = nxn-1; dv/dx = ex => ex
∫ exxn x dx = xnex - n∫exxn-1dx
In = xnex - nIn-1

∫ exx3 x dx = x3ex - 3∫exx2dx
= x3ex - 3[x2ex - 2∫exx dx]
=x3ex - 3x2ex + 6∫exx dx
=x3ex - 3x2ex + 6[xexx - ∫exx0dx]
=x3ex - 3x2ex + 6xexx - 6∫exdx
=x3ex - 3x2ex + 6xexx - 6ex + c

 

 

 

 

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