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## 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.

### 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 can be generalised for any product that is a composite function multiplied by the derivative of the inner funcion.

This integration rule follows from the chain rule for differentiation.

## Substitution Rule

To integrate f(g(x))g'(x) dx

1. Convert it to f(u) du by substituting u for g(x) and substituting du for g'(x) dx, taking out a common numeric factor, n, if present.
2. Evaluate f(u) du = F(u) + c or n f(u) du = nF(u) + c
3. Replace u by g(x) in F(u)

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.

### Examples

Evaluate Evaluate Evaluate ### Exercises

The hardest part of the integration is often choosing the right value to substitute. These exercises give practice in choosing u.

 What is the value of u to use to evaluate the indefinite integral: u =

These exercises are more tricky. Watch out for a common numeric factor in the derivative factor!

 What is the value of u and what is the value of the common numeric factor to use to evaluate the indefinite integral: u = common numeric factor =

Now try solving the whole problem. Evaluate the indefinite integrals.

 Evaluate the indefinite integral: =

## Quotients

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".

### Examples   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.

### Example The derivative of x2 + x − 1 is 2x + 1 not just 2x so this method cannot be used.

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