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Integration

Integration by Substitution 1

We assume that you are familiar with basic integration.

Recall that if d(F of x)by dx = f of x, then the indefinite integral the indefinite integral off(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, d(sine of (x squared)) by dx = cosine of (x squared)

Hence the indefinite integral of cos(x2)(2x) dx = sin(x2) + c (#)

If we substitute u = x2 then du by dx = 2x.

The left hand side of (#) becomes the indefinite integral of cos(u)(du by dx) dx, and the right hand side is sin(u) + c which equals the indefinite integral of cos(u) du.

Thus, the indefinite integral of cos(u)(du by dx) dx = the indefinite integral of 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 the indefinite integral of f(g(x))g'(x) dx

  1. Convert it to the indefinite integral of 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 the indefinite integral of f(u) du = F(u) + c or nthe indefinite integral of 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 integral of 4 times (4x plus 3) to the power 10 with respect to x = 1 eleventh times (4x plus 3) to the power 11 plus a constant

Evaluate integral of x times e to the power of x squared with respect to x = 1 half e to the power of x squared plus a constant

Evaluate integral of 14x over (x squared plus 1)  with respect to x = 7 times ln (x squared plus 1) plus a constant

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:

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:

indefinite integral

u = common numeric factor =

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

Evaluate the indefinite integral:

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:

the indefinite integral of = the indefinite integral of f(x)-1 f '(x) dx = ln | f(x) | + c

The derivative, f '(x), "disappears like magic".

Examples

eg1

eg3

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

eg4

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

Integration Index | Integration by Substitution 2 >>

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