<la-tex> Custom Element

Note

• The source code of this page contains <la-tex> custom elements with LaTeX expressions.
• This demo relies on MathML support to render various mathematical formulas.
• For alternative content, try this page explaining CSS transforms.

Basic Usage

The <la-tex> tag is a container for LaTeX expressions. The source is automatically converted into MathML using TeXZilla and the mathematical formula is displayed by your rendering engine. In particular, you need a standard-compliant HTML5 rendering engine as well as appropriate math fonts in order to get a good rendering.

You can use the <la-tex> tag to write inline mathematical expressions such as \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. The <la-tex> tag accepts the display and dir attributes of the math element. For instance you can write <la-tex display="block"> to insert display mathematical expressions such as

or <la-tex display="rtl"> to write right-to-left mathematical expressions such as س = \frac{-ب\pm\sqrt{ب^٢-٤اج}}{٢ا}. Note that the previous expression uses Arabic characters. In general you can use any Unicode characters in the LaTeX source: for example <la-tex>∑_{n=1}^{+∞} \frac{1}{n^2} = \frac{π^2}{6}</la-tex> will render as ∑_{n=1}^{+∞} \frac{1}{n^2} = \frac{π^2}{6}. If a parsing error occurs, then the original LaTeX source is displayed: \gamma + \frac{x}.

Just like the <math> tag, the <la-tex> tag can be used in other languages from the HTML5 family as shown below. Note that clicking on the HTML button below will retrieve the original LaTeX source, via the .textContent property of the <la-tex> element.

Finally, the <la-tex> tag will react to dynamic changes, as shown in the following example where we use Javascript to modify the display and dir attributes of the <la-tex> tag or to set its .textContent property. Inline Mode Display Mode ; LTR RTL ; Equation 1 Equation 2 Equation 3 Equation 4 \sqrt 2

Sample Document using <la-tex>

Copyright

      Copyright (C) 2013 Frédéric Wang

      Permission is granted to copy, distribute and/or modify this document
      under the terms of the GNU Free Documentation License, Version 1.3
      or any later version published by the Free Software Foundation;
      with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts.
      A copy of the license is included in the section entitled "GNU
      Free Documentation License".
    

Basic Integration

Ideas

If F is a primitive of f then \int_{a}^b f(x) d x = F(b) - F(a), provided the integrand and integral have no singularities on the path of integration. As a consequence, we can use a table of derivatives:

Examples

Using linearity and elementary power rule:

Using linearity and sine/cosine:

Using Leibniz rule, Chain rule and Exponential/Power:

Using the inverse function rule and tangent:

Using linearity, reciprocal rule and logarithm:

Using the power rules and chain rule:

Integration by Substitution

Idea

Do a substitution x = g(y) to simplify an integral. We have \frac{dx}{dy} = g'(y) and \int f(x) dx = \int f(g(y)) g'(y) dy: this is the chain rule!

Example 1

Let y = x^2 - 1, dy = 2 x dx then we get

We find the same result as

Example 2

Consider

Let y = x^3 - 2, dy = 3x^2 dx so

Example 3

Consider

Let u = -\frac{1}{v^2}, du = \frac{2}{v^3} dv:

Example 4

Consider

Let t = 2 \sin \theta and so dt = 2 \cos \theta d \theta. We have

We have -1 < \cos \theta < 0 on \left[\frac{3\pi}{4},\frac{7\pi}{6}\right] and so

Example 5

Consider

We first write

Now doing the substitution y = \frac{x - \frac{3}{2}}{\sqrt{\frac{5}{4}}}, dy = \frac{dx}{\sqrt{\frac{5}{4}}} we get

But we have already met this integral in the basic methods:

Example 6

Consider

We do the substitution t = \tan\left(\frac{\theta}{2}\right). We have d \theta = \frac{2 dt}{1+t^2}, \cos \theta = \frac{1-t^2}{1+t^2} and \sin \theta = \frac{2t}{1+t^2}. Also, we have

and so

and finally

Integration by parts

Idea

Given a two function f, g we have

this is Leibniz rule!

Example 1

Consider

We consider f(x) = x^2 + x + 1 and g(x) = \frac{e^{-3x}}{-3} that is f'(x) = 2x+1 and g'(x) = e^{-3x}. The integral can be written \int f g' dx and hence we get

We now consider h(x) = 2x+1, h'(x)=2 the second integral can be written \int h g' dx and hence

We now recognize g' in the last integral and so

Finally,

Example 2

We let f(x) = 2\ln x + 7\arctan x and g(x) = \frac{x^3}{3}. Hence f'(x) = \frac{2}{x} + \frac{7}{1+x^2} and g'(x) = x^3. The integration by parts gives:

The integral can be written

Finally

Example 3

Consider

We do an integration by parts with f(\theta) = \sin(3 \theta) and g'(\theta) = e^{2 \theta}:

We repeat a similar integration by parts on the remaining integral:

We note that we found the initial integral I. We get:

Finally,

Integration by Partial Fraction Decomposition

Idea

Given a rational function \frac{P}{Q}, write its partial fraction decomposition and integrate each term.

Example 1

Clearly, we have 6x^5-18x^4+18x^3-6x^2 = 6 x^2 (x^3-3x^2+3x-1) and 1 is a trivial root of the second factor so we can factor it by (x-1) and obtain:

Hence we search a decomposition of

for some constants A, B, C, D, E \in \mathbb{R}. We have

Finally,

Example 2

The denominator can be written u^5-4u^4+14u^3-20u^2+25u = u(u^4-4u^3+14u^2-20u+25). Evaluating the second factor at 0, \pm1, \pm2 we don't find any trivial roots. Hence we try to convert it to a depressed quartic by setting u = v - \frac{-4}{4} = v + 1. We find (u^4-4u^3+14u^2-20u+25) = v^4 + 8v^2 + 16 = (v^2 + 4)^2 = (u^2-2u+5)^2. The discriminant of u^2-2u+5 is 4(1-5) < 0 so we have the decomposition in irreducible factors:

We now try to find A, B, C, D, E \in \mathbb{R} such that

From the two last equalities, we immediately get D = 2 and E = -2. Finally,

We have

hence

Example 3

By polynomial long division we find that

and so

The integral of the first term is:

First we search trivial roots for the denominator. If we evaluate t^4+2t^3-t-2 at t=0,-1,1,-2,2 we find respectively -2, -2, 0, 0, 28. So we can factor X^4+2X^4-X-2 by (X - 1)(X + 2). We get

The polynomial X^2+X+1 is irreducible on \mathbb{R} because its discrimant is 1 - 4 = -3 < 0. Hence we search a decomposition of

for some constants A, B, C, D \in \mathbb{R}. We have

So

The integral of the two first terms is easy to compute:

The last term can be split into two parts:

We have

and

Finally we get

Alternatively, we can write

where t^\pm = \frac{-1\pm i \sqrt{3}}{2} are the complex roots of t^2+t+1. We have

We have

After simplification, we get

as we already found above.

Contour Integration

Idea

Consider contour integral \oint_\gamma f(z) d z along paths in complex plane to deduce integral along the real line.

Example 1

Consider the integral

Write z = e^{i \theta}, d z = i z d \theta. The integral becomes

where \gamma is the unit circle traversed counterclockwise. f : z \mapsto \frac{3(z^4+1)}{2 z^2 (z - 2 i)\left(z - \frac{i}{2}\right)} has two singularities inside that circle: 0 and \frac{i}{2}. Hence the integral is the 2 \pi i times the sum of the residues of f at these points:

Hence

Example 2

We want to know the value of the integral

For that purpose, we introduce the function

We note that x \mapsto \Im(f(x)) is the integrand above. We also note that the poles of f are z^\pm = -1 \pm 2 i. By choosing the right contour, one can show that

We have

Hence

Finally