Mastering Mathematics Smartly
by Wee Wen Shih

A unique self-help website that provides comprehensive coverage of mathematics at A-level & beyond, written in a student-friendly style.

Reduction formula

How is it possible for one to find \int_{}^{} sin^n x\: dx for n \geq 2?

In H2 mathematics, you have learnt the concepts of integration by parts and recurrences. We'll learn that these ideas will help us solve the above problem.

First, we rewrite the expression and then apply integration by parts, i.e.

\int_{}^{} sin^n x\: dx = \int_{}^{} (sin x)(sin^{n-1} x)\: dx
=(-cos x)(sin^{n-1} x) - \int_{}^{} (-cos x)\{(n-1)(sin^{n-2} x)(cos x)\:\} dx
=(-cos x)(sin^{n-1} x) + (n-1)\int_{}^{} (cos^2 x)(sin^{n-2} x)\: dx
=(-cos x)(sin^{n-1} x) + (n-1) \int_{}^{} sin^{n-2} x\:dx - (n-1) \int_{}^{} sin^n x\:dx , since cos^2 x = 1 - sin^2 x



Let I_n = \int_{}^{}sin^n x\: dx. Then

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

\Rightarrow nI_n =(-cos x)(sin^{n-1} x) + (n-1)I_{n-2}

\Rightarrow I_n =\frac{1}{n}[(-cos x)(sin^{n-1} x) + (n-1)I_{n-2}]

So, in order for us to find I_n, it is necessary to find I_{n-2}, I_{n-4}, ... i.e. we make use of the approach similar to recurrences.

For example, I_4 = \frac{1}{4}[(-cos x)(sin^3 x) + 3I_2] and I_2 = \frac{1}{2}[(-cos x)(sin x) + 3I_0] and I_0 = \int_{}^{}1\: dx.

Practice: Let I_n=\int_{0}^{\frac{\pi}{2}} x^n cos x\: dx. Show that I_n = \left(\frac{\pi}{2}\right)^n - n(n-1)I_{n-2}, n \geq 2. Hence find I_6 exactly.

Stationary values

In H2 mathematics, you learnt how to find the stationary points of any function involving the only variable x.

We will find out soon the approach to finding the stationary points of a function involving two variables x and y, by means of partial derivatives.

Suppose we wish to find the stationary points of f(x, y) = x^3 - 3x^2y + y^2 + 4y.

The first step is to find the partial derivatives \frac{\partial f}{\partial x} and \frac{\partial f}{\partial y}. To find \frac{\partial f}{\partial x}, we differentiate f with respect to x while keeping y contant. In the same fashion, we differentiate f with respect to y while holding x contant in order to find \frac{\partial f}{\partial y}.

For our example above, we obtain \frac{\partial f}{\partial x} = 3x^2 - 6xy and \frac{\partial f}{\partial y} = -3x^2 + 2y + 4.

The second step is to find x, y values when \frac{\partial f}{\partial x} = 0 and \frac{\partial f}{\partial y} = 0.

So 3x^2 - 6xy = 0 -- (1)
and
-3x^2 + 2y + 4 = 0 -- (2)

From (1): 3x(x - 2y) = 0 \Rightarrow x = 0 \:or\: x = 2y

Substitute x = 0 into (2): y = -2

Substitute x = 2y into (2): -12y^2 + 2y + 4 = 0

\Rightarrow 6y^2 - y - 2 = 0

\Rightarrow (2y + 1)(3y - 2) = 0

\Rightarrow y = -\frac{1}{2} \:or\: y = \frac{2}{3}

Hence, the stationary points are \left(0, -2 \right), \left(-1, -\frac{1}{2} \right) and \left(\frac{4}{3}, \frac{2}{3} \right).

Practice: Find the stationary points of f(x, y) = x^2 + y^3 - 3xy.

Tangent planes

From H2 Mathematics, we know that in a two-dimensional x-y plane, we can visualise the line tangent to a curve at a point.

In the case of a three-dimensional x-y-z plane, we can picture the plane tangent to a surface at a point.

Let us study an example: Consider the surface given by z = 5x^2 + 3y + xy. Find the equation of the tangent plane at (1, 2).

We will need to carry out partial derivatives and the method is given below.

In general, the equation of the tangent plane at (x_0, y_0, z_0) to a surface given by z = \text{f}(x, y) is:

z_x(x_0, y_0)\:(x - x_0) + z_y(x_0, y_0)\:(y - y_0) - (z - z_0) = 0, where z_0 = \text{f}(x_0, y_0).
 
This demonstration helps us to visualise the tangent to a surface.

As an exercise, apply the four steps for this practice question:

Consider the surface given by z = x + \frac{x}{y + 1}. Find the equation of the tangent plane at (2, 1).