Integration by Reduction Formulae

Suppose you have to ∫e^x sin(x)dx.

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

u = e^x; dv/dx = sin x
So, du/dx = e^x; v = -cos x
∫e^x sin(x)dx = -e^xcos x + ∫ e^xcos x dx 1
Now, we have to repeat the integration process for ∫ e^xcos x dx, which is as follows:
u = e^x; dv/dx = cos x
du/dx = e^x; v = sin x
So, ∫ e^xcos x dx = e^xsin x - ∫ e^xsin 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 ∫ e^xcos x dx the subject from both equations and equate them.
From 1 => ∫e^x sin(x)dx + e^xcos x = ∫ e^xcos x dx 3
From 2 => ∫ e^xcos x dx = e^xsin x - ∫ e^xsin x dx 4
Since 3 = 4 => ∫e^x sin(x)dx + e^xcos x = e^xsin x - ∫ e^xsin x dx
2∫e^x sin(x)dx = e^xsin x - e^xcos x
∫e^x sin(x)dx = 1/2e^x[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 something 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 down 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 = I_n, then ∫[ln x]^n-1 = I_n-1
Therefore,
I_n = x [ln x]ⁿ - n I_n-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 = ∫sin^n-1x sin x dx
Let u = ∫sin^n-1x and dv/dx = sin x
So, du/dx = (n-1)sin^n-2x cos x; v = -cos x
∫sinⁿx dx = -sin^n-1 x cos x + ∫ cos x (n-1)sin^n-2x cos x dx
= -sin^n-1x cos x + ∫ cos² x (n-1)sin^n-2x dx
= -sin^n-1x cos x + ∫ [1 - sin²x ](n-1)sin^n-2x dx
= -sin^n-1x cos x + ∫ (n-1)sin^n-2x dx - ∫(n-1)sin²x sin^n-2x dx
= -sin^n-1x cos x + ∫ (n-1)sin^n-2x dx - ∫(n-1)sinⁿx dx
∫n sinⁿx dx = -sin^n-1x cos x + ∫ (n-1)sin^n-2x dx
∫sinⁿx dx = -1/n sin^n-1x cos x+ ∫ (n-1)/n sin^n-2x dx
I_n = -1/n sin^n-1 x cos x + (n-1)/n I_n-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 = ∫cos^n-1x cos x dx
Let u = ∫cos^n-1x and dv/dx = cos x
So, du/dx = -(n-1)cos^n-2x sin x; v = sin x
∫cosⁿx dx = -cos^n-1 x sin x + ∫ sin x (n-1)cos^n-2x sin x dx
= cos^n-1x sin x + ∫ sin² x (n-1)cos^n-2x dx
= cos^n-1x sin x + ∫ [1 - cos²x ](n-1)cos^n-2x dx
= cos^n-1x sin x + ∫ (n-1)cos^n-2x dx - ∫(n-1)cosⁿx dx
∫n cosⁿx dx = cos^n-1x sin x + ∫(n-1)cos^n-2x dx
∫cosⁿx dx = 1/n cos^n-1x sin x + ∫ (n-1)/n cos^n-2x dx
I_n = 1/n cos^n-1 x sin x + (n-1)/n I_n-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 = ∫tan^n-2x tan²x dx
u = tan^n-2x => du/dx = n-2 tan^n-3x sec²x; dv/dx tan²x = sec²x - 1 => v = tan x - x
The integration leads to the following formula:
∫tanⁿ x dx = 1/n-1 tan^n-1 x - ∫tan^n-2 x dx
I_n = 1/n-1 I_n-1 - I_n-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 ∫ e^xxⁿ x dx and hence find ∫e^xx³dx.

u = xⁿ => du/dx = nx^n-1; dv/dx = e^x => e^x
∫ e^xxⁿ x dx = xⁿe^x - n∫e^xx^n-1dx
I_n = xⁿe^x - nI_n-1

∫ e^xx³ x dx = x³e^x - 3∫e^xx²dx
= x³e^x - 3[x²e^x - 2∫e^xx dx]
=x³e^x - 3x²e^x + 6∫e^xx dx
=x³e^x - 3x²e^x + 6[xe^xx - ∫e^xx⁰dx]
=x³e^x - 3x²e^x + 6xe^xx - 6∫e^xdx
=x³e^x - 3x²e^x + 6xe^xx - 6e^x + c

You can practise the graphical representations of the above integrals interactively here. Please choose one example at a time to stop the graphics from being blurred.

Reduction Formula - interactive graphical practice

This is an awesome opportunity for you to practise the integration by reduction formulae; it's fully interactive and you need to take the following steps very carefully for it to to work.
Write the formula inside the text box: in order to raise to the power, please use, ^, of your keyboard, followed by the power. For example, sin²(x) has to be entered as sin^2(x). Then press enter.
The blue curve is the function to be integrated; the red curve is the integrated function.