A tutorial, with examples and detailed solutions, in using the properties of indefinite integrals in calculus is presented. A set of exercises with answers is presented after the tutorial.

In what follows, c is a constant of integration and can take any contant value.

Let u = x^{ 2} - 1, du/dx = 2x and the given integral can be written as

(x^{ 2} - 1)^{ 20} 2x dx

= u^{ 20} (du/dx) dx

= u^{ 20} du according to above property

= u^{ 21} / 21 + c = (x^{ 2} - 1)^{ 21} / 21 + c

3 - Integration by Parts.

f(x) g '(x) dx = f(x) g(x) - f '(x) g(x) dx

Example: Evaluate the integral

x cos x dx

Solution:

Let f(x) = x and g ' (x) = cos x which gives

f ' (x) = 1 and g(x) = sin x

From integration by parts formula above,

x cos x dx = x sin x - 1 sin x dx

= x sin x + cos x + c

Exercises: Use the table of integrals and the properties above to evaluate the following integrals. [Note that you may need to use more than one of the above properties for one integral].

1. (1/2) ln x dx

2. [sin x + x^{ 5}] dx

3. [sinh x - 3] dx

4. -x sin x dx

Answers to Above Exercises

1. [x ln x] / 2 - x / 2

2. x^{ 6} / 6 - cos x

3. cosh x - 3x

4. x cos x - sin x

More references on
integrals and their applications in calculus.