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.