# Primitive function/Substitution/Section

Suppose that denotes a real interval, and let

denote a continuous function. Let

be a continuously differentiable function. Then

holds.

Since is continuous and is continuously differentiable, both integrals exist. Let denote a primitive function for , which exists, due to fact. Because of the chain rule, the composite function

has the derivative . Therefore,

Typical examples, where one sees immediately that one can apply substitution, are

with the primitive function

or

with the primitive function

Often, the indefinite integral is not in a form where one can apply the preceding rule directly. Then the following variant is more appropriate.

Suppose that

is a continuous function, and let

denote a bijective continuously differentiable function. Then

holds.

Because of fact, we have

The substitution is applied in the following way: suppose that the integral

has to be computed. Then one needs an idea that the integral gets simpler by the substitution

(taking into account the derivative and that the inverse function has to be determined). Setting and , we have the situation

In certain cases, some standard substitutions help.

In order to make a substitution, three operations have to be done.

- Replace by .
- Replace by .
- Replace the integration bounds and by and .

To remember the second step, think of

which in the framework "differential forms“, has a meaning.

The upper curve of the unit circle is the set

For a given , , there exists exactly one fulfilling this condition, namely . Hence, the area of the upper half of the unit circle is the area beneath the graph of the function , above the interval , that is

Applying substitution with

(where is bijective, due to fact), we obtain, using example, the identities

In particular, we get that

is a primitive function for . Therefore,