Rules of Integrals with Examples

A tutorial, with examples and detailed solutions, in using the rules of indefinite integrals in calculus is presented. A set of questions with solutions is also included.

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


1 - Integral of a power function: \( f(x) = x^n \)

Integral of a Power x^n
Example: Evaluate the integral \[ \int x^5 dx \]
Solution: \[ \int x^5 dx = \dfrac{x^{5 + 1}}{ 5 + 1} + c = \dfrac{x^6}{6} + c \]


2 - Integral of a function \( f \) multiplied by a constant \( k \): \( k f(x) \)

Integral of the Multiple of a function
Example: Evaluate the integral \[ \int 5 \sin \; x dx \] Solution:
According to the above rule
\( \displaystyle \int 5 \sin (x) dx = 5 \int \sin(x) dx \)
\( \displaystyle \int \sin(x) dx \)   is given by 2.1 in table of integral formulas, hence
Hence
\( \displaystyle \int 5 \sin(x) dx = - 5 \cos x + C \)


3 - Integral of Sum of Functions.

Integral of the Sum of functions
Example: Evaluate the integral \[ \int (x + e^x) dx \]
Solution:
According to the above property
\( \displaystyle \int (x + e^x) dx = \int x \; dx + \int e^x \; dx \)
\( \displaystyle \int x \; dx \) is given by 1.3 and \( \displaystyle \int e^x \; dx \) by 4.1 in table of integral formulas, hence
\[ \int (x + e^x) \; dx = \dfrac{x^2}{2}x + e^x + c \]


4 - Integral of Difference of Functions.

Integral of the Difference of two functions
Example: Evaluate the integral \[ \int (2 - 1/x) \; dx \]
Solution:
According to the above property
\( \displaystyle \int (2 - 1/x) dx = \int 2 \; dx - \int (1/x) \; dx \)
\( \int 2 \; dx \) is given by 1.2 and \( \int (1/x) \; dx \) by 1.4 in table of integral formulas, hence
\[ \int (2 - 1/x) \; dx = 2x - \ln |x| + c \]


5 - Integration by Substitution.

Integral by Substitution
Example: Evaluate the integral \[ \int (x^2 - 1)^{20} 2x \; dx \]
Solution:
Let \( u = x^2 - 1\), hence \( du/dx = 2x \) and the given integral can be written as
\( \displaystyle \int(x^2 - 1)^{20} \; 2x \; dx = \int u^{20} (du/dx) dx = \int u^{20} du \)
which evaluates to
\( = \dfrac{u^{21}}{21} + c \)
Substitute back
\( = \dfrac{(x^2 - 1)^{21}}{21} + c \)


6 - Integration by Parts.

Integral by Parts
Example: Evaluate the integral \[ \int \; x \; \cos x \; dx \]
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,
\( \displaystyle \int \; x \cos x \; dx = x sin x - \int 1 \sin x dx \)
\( = x \sin x + \cos x + c \)

More Questions with Solutions

Use the table of integral formulas and the rules above to evaluate the following integrals. [Note that you may need to use more than one of the above rules for one integral].
1. \( \displaystyle \int (1 / 2) \ln \; (x) dx \)
2. \( \displaystyle \int (\sin (x) + x^5 ) \; dx \)
3. \( \displaystyle \int (\sinh (x) - 3) \; dx \)
4. \( \displaystyle \int - x \sin (x) \; dx \)
5. \( \displaystyle \int \sin^{10}(x) \; \cos(x) dx \)

Solutions to the Above Questions


1.
This is the integral of ln (x) multiplied by 1 / 2 and we therefore use rule 2 above to obtain:
\( \int \)(1 / 2) ln (x) dx = (1 / 2) \( \int \)ln (x) dx
We now use formula 4.3 in the table of integral formulas to evaluate \( \int \)ln (x) dx. Hence
\( \int \)(1 / 2) ln (x) dx = (1 / 2) ( (x ln (x)) - x ) + c

2.
Use rule 3 ( integral of a sum ) to obtain
\( \int \)[sin (x) + x 5] dx = \( \int \) sin (x) dx + \( \int \)x 5 dx
We use formula 2.1 in the table of integral formulas to evaluate \( \int \) sin (x) dx and rule 1 above to evaluate \( \int \)x 5 dx. Hence
\( \int \)[sin (x) + x 5] dx = - cos (x) + x 6 / 6
3.
Use rule 4 (integral of a difference) to obtain
\( \int \)(sinh (x) - 3) dx = \( \int \) sinh (x) dx - \( \int \)3 dx
We use formula 7.1 in the table of integral formulas to evaluate \( \int \) sinh (x) dx and integral of the constant 3 to obtain
\( \int \)(sinh (x) - 3) dx = cosh (x) - 3 x + c
4.
The integrand is the product of two function x and sin (x) and we try to use integration by parts in rule 6 as follows:
Let f(x) = x , g'(x) = sin(x) and therefore g(x) = - cos(x)
Hence
\( \int \) - x sin (x) dx = - \( \int \) f(x) g'(x) dx = - ( f(x) g(x) - \( \int \) f'(x) g(x) dx)
Substitute f(x), f'(x), g(x) and g'(x) by x , 1, sin(x) and - cos(x) respectively to write the integral as
= - x (- cos(x)) + \( \int \) 1 (- cos(x)) dx
Use formula 2.2 in in the table of integral formulas to evaluate \( \int \) cos(x) dx and simplify to obtain
= x cos (x) - sin(x) + c

5.
Let u = sin(x) and therefore du/dx = cos(x). Hence the given integral can be written as
\( \int \) sin10(x) cos dx = \( \int \)( u10 du/dx ) dx
Use rule 5 to write
= \( \int \) u10 du
which gives
= u 11 / 11 + c
Substitute u by sin(x) to obtain
= (1 / 11) (sin 11(x) ) + c

More References and Links

Table of Integral Formulas
integrals and their applications in calculus.
evaluate integrals.
Integration by Substitution.