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In what follows, c is a constant of integration and can take any contant value.
 5 sinx dxSolution: According to the above property 5 sinx dx = 5sin(x) dx
sin(x) dx is given by 2.1 in table of integrals, hence
5 sin(x) dx = -5cos x + c
2 - Integral of Sum of Functions.[f(x) + g(x)] dx = f(x) dx + g(x) dx
Example: Evaluate the integral [x + e x] dxSolution: According to the above property [x + e x] dx = x dx + e x dx
ò x dx is given by 1.3 and e x dx
by 4.1 in table of integrals, hence
[x + e x] dx = x 2 / 2 + e x + c
3 - Integral of Difference of Functions.[f(x) - g(x)] dx = f(x) dx - g(x) dx
Example: Evaluate the integral [2 - 1/x] dxSolution: According to the above property [2 - 1/x] dx = 2 dx - (1/x) dx
2 dx is given by 1.2 and (1/x) dx by 1.4 in table of integrals, hence
[2 - 1/x] dx = 2x - ln |x| + c
3 - Integration by Substitution.[f(u) du/dx] dx = f(u) du
Example: Evaluate the integral (x 2 - 1) 20 2x dxSolution: 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 dxSolution: 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].
More references on integrals and their applications in calculus. |
