Integration by Substitution 1
We assume that you are familiar with basic integration.
Recall that if , then the indefinite integral f(x) dx = F(x) + c.
Note that there are no general integration rules for products and quotients of two functions.
We now provide a rule that can be used to integrate products and quotients in particular forms. The idea is to convert an integral into a basic one by substitution.
Before we give a general expression, we look at an example.
By the chain rule,
Hence cos(x2)(2x) dx = sin(x2) + c…… (#)
If we substitute u = x2 then = 2x.
The left hand side of (#) becomes cos(u)() dx, and the right hand side is sin(u) + c which equals cos(u) du.
Thus, cos(u)() dx = cos(u) du , where u = x2.
This integration rule follows from the chain rule for differentiation.
To integrate f(g(x))g'(x) dx
NB. The presence of the derivative as a factor of what is being substituted into an integrand is an essential ingredient of the substitution rule.
The hardest part of the integration is often choosing the right value to substitute. These exercises give practice in choosing u.
These exercises are more tricky. Watch out for a common numeric factor in the derivative factor!
Now try solving the whole problem. Evaluate the indefinite integrals.
One type of problem deserves special mention. The problem appears at first sight as a quotient with larger denominator than numerator rather than as a product of a function raised to the power -1 and its derivative and thus often fools solvers when viewed in isolation:
= f(x)-1 f '(x) dx = ln | f(x) | + c
The derivative, f '(x), "disappears like magic".
Be very careful to check that the numerator is equal to a constant times the derivative of the denominator or this method cannot be used.
The derivative of x2 + x − 1 is 2x + 1 not just 2x so this method cannot be used.