## Integration Techniques

Many integration formulas can be derived directly from their corresponding derivative formulas, while other integration problems require more work. Some that require more work are substitution and change of variables, integration by parts, trigonometric integrals, and trigonometric substitutions.

## Basic formulas

Most of the following basic formulas directly follow the differentiation rules.

Example 1: Evaluate

Using formula (4) from the preceding list, you find that .

Example 2: Evaluate .

Because using formula (4) from the preceding list yields

Example 3: Evaluate

Applying formulas (1), (2), (3), and (4), you find that

Example 4: Evaluate

Using formula (13), you find that

Example 5: Evaluate

Using formula (19) with a = 5, you find that

## Substitution and change of variables

One of the integration techniques that is useful in evaluating indefinite integrals that do not seem to fit the basic formulas is substitution and change of variables. This technique is often compared to the chain rule for differentiation because they both apply to composite functions. In this method, the inside function of the composition is usually replaced by a single variable (often u). Note that the derivative or a constant multiple of the derivative of the inside function must be a factor of the integrand.

The purpose in using the substitution technique is to rewrite the integration problem in terms of the new variable so that one or more of the basic integration formulas can then be applied. Although this approach may seem like more work initially, it will eventually make the indefinite integral much easier to evaluate.

Note that for the final answer to make sense, it must be written in terms of the original variable of integration.

Example 6: Evaluate

Because the inside function of the composition is x 3 + 1, substitute with

Example 7:

Because the inside function of the composition is 5 x, substitute with

Example 8: Evaluate

Because the inside function of the composition is 9 – x 2, substitute with

## Integration by parts

Another integration technique to consider in evaluating indefinite integrals that do not fit the basic formulas is integration by parts. You may consider this method when the integrand is a single transcendental function or a product of an algebraic function and a transcendental function. The basic formula for integration by parts is

where u and v are differential functions of the variable of integration.

A general rule of thumb to follow is to first choose dv as the most complicated part of the integrand that can be easily integrated to find v. The u function will be the remaining part of the integrand that will be differentiated to find du. The goal of this technique is to find an integral, ∫ v du, which is easier to evaluate than the original integral.

Example 9: Evaluate ∫ x sec 2 x dx.

Example 10: Evaluate ∫ x 4 In x dx.

Example 11: Evaluate ∫ arctan x dx.

Integrals involving powers of the trigonometric functions must often be manipulated to get them into a form in which the basic integration formulas can be applied. It is extremely important for you to be familiar with the basic trigonometric identities, because you often used these to rewrite the integrand in a more workable form. As in integration by parts, the goal is to find an integral that is easier to evaluate than the original integral.

Example 12: Evaluate ∫ cos 3 x sin 4 x dx

Example 13: Evaluate ∫ sec 6 x dx

Example 14: Evaluate ∫ sin 4 x dx

If an integrand contains a radical expression of the form a specific trigonometric substitution may be helpful in evaluating the indefinite integral. Some general rules to follow are

1. If the integrand contains

2. If the integrand contains

3. If the integrand contains

Right triangles may be used in each of the three preceding cases to determine the expression for any of the six trigonometric functions that appear in the evaluation of the indefinite integral.

Example 15: Evaluate

Because the radical has the form

Figure 1 Diagram for Example 15.

Example 16: Evaluate

Because the radical has the form

Figure 2 Diagram for Example 16.