Mathematical principles of indefinite integrals are presented. This course starts with the definition of integral and its main properties. Rules of both integration by substitution and integration by parts are shown. Study of given examples can help to train analytic techniques of integration.
G. M. Fichtenholz – Integral and differential calculus, vol. 2, PWN Warsaw 1999 (originally in Russian)
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The course is divided into paragraphs. The first part of tutorial presents the elementary idea of finding indefinite integrals together with basic examples (as shown below). The two main ways of integration i. e. integration by substitution and integration by parts are considered. Each main part of handbook follows brief theoretical introduction (e. g. Indefinite integrals. Definition.). This course explains simple and more complicated cases of integration. Among the advanced cases are: integration of rational functions, integration of functions with roots and integration of trigonometric functions. In these examples transformations must be applied to starting expressions to obtain better form of integrated functions that, in turn, can be easily solved.
Examples presented-below show the way of putting forward solutions. Each essential part of finding the output is treated separately. Thus one can make use of only part of hints and after finding the solution compare this result with the outcome presented in the tutorial.
Let's consider a simple integral of a trigonometric function.
As the first step we transform the trigonometric function to a more convenient form:
After that finding the solution is trivial:
|∫||dx/cos2(x) -||∫||dx = tg(x) - x + C|
The second example is an example of integration by parts.
by parts transformation:
and the solution reads:
|x2/2 - x/4 sin(2x) - 1/2||∫||xdx + 1/4||∫||sin(2x)dx =|
|x2/2 - x/4 sin(2x) - x2/4 - 1/8 cos(2x) + C =||x2/4 - x/4 sin(2x) - 1/8 cos(2x) + C|
Another example shows the main idea of integration by substitution.
|∫||xdx/[1 - x2]1/2|
the obvious way of finding solution is to make the substitution:
the solution is:
|-1/2||∫||dt/t1/2 = -t1/2 + C = -(1 - x2)1/2 + C|
Integration by substitution - second example, trigonometric function.
the solution is:
|∫||-sin(t)dt = cos(t) + C = cos(1/x) + C|