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^{x}cos x + ∫ e^{x}cos x dx 1

Now, we have to repeat the integration process for ∫ e^{x}cos x dx, which is as follows:

u = e^{x}; dv/dx = cos x

du/dx = e^{x}; v = sin x

So, ∫ e^{x}cos x dx = e^{x}sin x - ∫ e^{x}sin 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^{x}cos x dx the subject from both equations and equate them.

From 1 => ∫e^{x} sin(x)dx + e^{x}cos x = ∫ e^{x}cos x dx 3

From 2 => ∫ e^{x}cos x dx = e^{x}sin x - ∫ e^{x}sin x dx 4

Since 3 = 4 => ∫e^{x} sin(x)dx + e^{x}cos x = e^{x}sin x - ∫ e^{x}sin x dx

2∫e^{x} sin(x)dx = e^{x}sin x - e^{x}cos 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 easy to come up with. **Integration by Reduction Formulae** is one such method.

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.

∫ [ln x]^{n}dx

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]^{n}dx = 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 I_{3}.

I_{3} = x[ln x]^{3} - 3 I_{2}]

= x(ln x)^{3} - 3 [x (ln x)^{2} - 2 I_{1}]

= x(ln x)^{3} - 3x (ln x)^{2} + 6 I_{1}

= x(ln x)^{3} - 3x (ln x)^{2} + 6 [x (ln x)^{1} - I_{0}]

= x(ln x)^{3} - 3x (ln x)^{2} + 6x (ln x)^{1} - 6I_{0}

= 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 ∫sin^{n} x dx and hence find ∫sin^{4} x dx.

∫sin^{n}x dx = ∫sin^{n-1}x sin x dx

Let u = ∫sin^{n-1}x and dv/dx = sin x

So, du/dx = (n-1)sin^{n-2}x cos x; v = -cos x

∫sin^{n}x dx = -sin^{n-1} x cos x + ∫ cos x (n-1)sin^{n-2}x cos x dx

= -sin^{n-1}x cos x + ∫ cos^{2} x (n-1)sin^{n-2}x dx

= -sin^{n-1}x cos x + ∫ [1 - sin^{2}x ](n-1)sin^{n-2}x dx

= -sin^{n-1}x cos x + ∫ (n-1)sin^{n-2}x dx - ∫(n-1)sin^{2}x sin^{n-2}x dx

= -sin^{n-1}x cos x + ∫ (n-1)sin^{n-2}x dx - ∫(n-1)sin^{n}x dx

∫n sin^{n}x dx = -sin^{n-1}x cos x + ∫ (n-1)sin^{n-2}x dx

∫sin^{n}x dx = -1/n sin^{n-1}x cos x+ ∫ (n-1)/n sin^{n-2}x dx

I_{n} = -1/n sin^{n-1} x cos x + (n-1)/n I_{n-2}

∫ sin^{4}x dx = -1/4 sin^{3}x cos x + ∫ 3/4 sin^{2}x dx

= -1/4 sin^{3}x cos x + 3/4[-1/2 cos x sin x + ∫ 1/2 sin^{0}x dx]

= -1/4 sin^{3}x cos x -3/8 cos x sin x + 3/8 x + c dx

= -1/4 sin^{3}x cos x -3/8 sin x cos x + 3/8 x + c

since sin 2x = 2 sin x cos x

∫ sin^{4}x dx = -1/4 sin^{3}x cos x - 3/16 sin 2x + 3/8 x + c

**E.g.2**

Find a reduction formula to integrate ∫cos^{n} x dx and hence find ∫cos^{4} x dx.

∫cos^{n}x dx = ∫cos^{n-1}x cos x dx

Let u = ∫cos^{n-1}x and dv/dx = cos x

So, du/dx = -(n-1)cos^{n-2}x sin x; v = sin x

∫cos^{n}x dx = -cos^{n-1} x sin x + ∫ sin x (n-1)cos^{n-2}x sin x dx

= cos^{n-1}x sin x + ∫ sin^{2} x (n-1)cos^{n-2}x dx

= cos^{n-1}x sin x + ∫ [1 - cos^{2}x ](n-1)cos^{n-2}x dx

= cos^{n-1}x sin x + ∫ (n-1)cos^{n-2}x dx - ∫(n-1)cos^{n}x dx

∫n cos^{n}x dx = cos^{n-1}x sin x + ∫(n-1)cos^{n-2}x dx

∫cos^{n}x dx = 1/n cos^{n-1}x sin x + ∫ (n-1)/n cos^{n-2}x dx

I_{n} = 1/n cos^{n-1} x sin x + (n-1)/n I_{n-2}

∫ cos^{4}x dx = 1/4 cos^{3}x sin x + ∫ 3/4 cos^{2}x dx

= 1/4 cos^{3}x sin x + 3/4[1/2 cos x sin x + ∫ 1/2 cos^{0}x dx]

= 1/4 cos^{3}x sin x + 3/8 cos x sin x + 3/8 x + c

Since sin 2x = 2 sin x cos x

∫ cos^{4}x dx = 1/4 cos^{3}x sin x + 3/16 sin 2x + 3/8 x + c

**E.g.3**

Find a reduction formula to integrate ∫ tan^{n} x dx and hence find ∫tan^{4}x^{ dx.}

∫tan^{n} x dx = ∫tan^{n-2}x tan^{2}x dx

u = tan^{n-2}x => du/dx = n-2 tan^{n-3}x sec^{2}x; dv/dx tan^{2}x = sec^{2}x - 1 => v = tan x - x

The integration leads to the following formula:

∫tan^{n} 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}

∫tan^{4} x dx = 1/3 tan^{3} x - ∫tan^{2} x dx

= 1/3 tan^{3} x - [tan x - ∫ dx]

= 1/3 tan^{3} x - tan x + x + c

**E.g.4**

Find a reduction formula to integrate ∫ e^{x}x^{n} x dx and hence find ∫e^{x}x^{3}dx.

u = x^{n} => du/dx = nx^{n-1}; dv/dx = e^{x} => e^{x}

∫ e^{x}x^{n} x dx = x^{n}e^{x} - n∫e^{x}x^{n-1}dx

I_{n} = x^{n}e^{x} - nI_{n-1}

∫ e^{x}x^{3} x dx = x^{3}e^{x} - 3∫e^{x}x^{2}dx

= x^{3}e^{x} - 3[x^{2}e^{x} - 2∫e^{x}x dx]

=x^{3}e^{x} - 3x^{2}e^{x} + 6∫e^{x}x dx

=x^{3}e^{x} - 3x^{2}e^{x} + 6[xe^{x}x - ∫e^{x}x^{0}dx]

=x^{3}e^{x} - 3x^{2}e^{x} + 6xe^{x}x - 6∫e^{x}dx

=x^{3}e^{x} - 3x^{2}e^{x} + 6xe^{x}x - 6e^{x} + c

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