To do integration by substitution using Leibniz notation, we think of the derivative function as a fraction of infinitesimally small quantities du and dx. We change variables by manipulating these infinitesimal quantities.
The general strategy is pretty much the same as before:
- Change variables (substitute in u for some function of x).
- Integrate.
- Put the original variable back (substitute the function of x back in for u).
Example 1
Find . |
Example 2
|
Example 3
For the integral, (a) identify u and du and (b) integrate by substitution. |
Example 4
For the integral, (a) identify u and du and (b) integrate by substitution. |
Example 5
For the integral, (a) identify u and du and (b) integrate by substitution. |
Example 6
Find . |
Example 7
Find . |
Example 8
Example. Find without using the "multiplying by 1" trick. |
Exercise 1
For the integral, (a) identify u and du and (b) integrate by substitution.
Exercise 2
For the integral, (a) identify u and du and (b) integrate by substitution.
Exercise 3
For the integral, (a) identify u and du and (b) integrate by substitution.
Exercise 4
For the integral, (a) identify u and du and (b) integrate by substitution.
Exercise 5
For the integral, (a) identify u and du and (b) integrate by substitution.
Exercise 6
Integrate by substitution.
Exercise 7
Integrate by substitution.
Exercise 8
Integrate by substitution.
Exercise 9
Integrate by substitution.
Exercise 10
Integrate by substitution.
Exercise 11
Integrate.
Exercise 12
Integrate.
Exercise 13
Integrate.
Exercise 14
Integrate.
Exercise 15
Integrate.