Other Integration Methods

When evaluating integrals in calculus, there are several integration methods to keep in mind. More complicated integration problems might be solvable using a method called “integration by parts”.

In this section we discuss some other integration methods such as u-substitution , trigonometric integration and partial fractions methods.

In the u-substitution method, the goal is to rewrite a complicated integral as a simpler integral that can be evaluated using the integration power rule.

For the trigonometric integration method, the goal is to convert algebraic expressions that we may not be able to integrate into expressions involving trigonometric functions such as sin(x) and cos(x), which we may be able to integrate.

The method called “partial fractions” is useful when the integrand contains a rational expression. Then the integrand can be rewritten as the sum of fractions and then each fraction is integrated separately.

The video below provides more details and explanations for these various integration methods together with several worked-out examples.

Background:

When evaluating integrals in calculus, there are several integration methods that can be used. For simpler integrals the power rule is often applied. in mind. More complicated integration problems might be solvable using a method called “integration by parts”.

In this section we discuss some other integration methods such as u-substitution method, trigonometric integration and partial fractions method.

U-substitution method

In the u-substitution method, the goal is to rewrite a complicated integral as a simpler integral that can be evaluated using the integration power rule.

The u-substitution method is often described as the “reverse chain rule” because it undoes the process of differentiating a composite function. The goal is to replace a complicated part of the integrand with a single variable u, which makes the integral easier to evaluate. The resulting simplified integral can often be solved using the power rule from integration.

Typically, we choose u as an inner function whose derivative also appears (or nearly appears) in the integrand. Once the substitution is made, we also replace dx with du using the relationship \( du = f'(x)\,dx \). After rewriting, the integral in terms of u is usually simpler to solve. Finally, after integrating with respect to u, we substitute back the original expression for u to get the solution in terms of x. This method is especially useful for integrals involving powers of polynomial expressions, trigonometric functions, or exponential expressions.

Example:
Evaluate the integral:

\( \int 4x^3 (x^4 + 5)^7 \, dx \)

Step 1. Choose substitution:

Let \( u = x^4 + 5 \), so that \( du = 4x^3 \, dx \).

Step 2. Rewrite the integral:

\( \int (x^4 + 5)^7 (4x^3 \, dx) = \int u^7 \, du \)

Step 3. Integrate with respect to u:

\( \int u^7 \, du = \frac{u^8}{8} + C \)

Step 4. Substitute back:

\( = \frac{(x^4 + 5)^8}{8} + C \)

Final Answer:

\( \int 4x^3 (x^4 + 5)^7 \, dx = \frac{(x^4 + 5)^8}{8} + C \)

Trigonometric Integration

This method often applies when integrals include powers of sine, cosine, tangent, or other trigonometric expressions. The strategy depends on the form of the integral:

(1) when the integrand contains odd or even powers of sine or cosine, trigonometric identities are used to reduce powers;
(2) when tangent and secant are involved, substitutions such as \( u = \tan(x) \) or \( u = \sec(x) \) may be useful;
(3) for products of trigonometric functions, identities like \( \sin^2(x) + \cos^2(x) = 1 \), or half-angle formulas, can simplify the expression.

By rewriting the integral in terms of simpler functions, one can apply basic integration rules or substitution. This method is widely applied in solving integrals in physics, engineering, and mathematics. Careful selection of identities and substitutions is a key aspect for this trigonometric integration method.

Example:
Evaluate the integral:

\( \int \sin^4(x) \cos(x) \, dx \)

Step 1: Choose substitution

Let \( u = \sin(x) \), so that \( du = \cos(x) \, dx \).

Step 2. Rewrite the integral:

\( \int \sin^4(x) \cos(x) \, dx = \int u^4 \, du \)

Step 3. Integrate with respect to u:

\( \int u^4 \, du = \frac{u^5}{5} + C \)

Step 4. Substitute back:

Substituting back \( u = \sin(x) \):

\( \frac{\sin^5(x)}{5} + C \)

Final Answer:

\( \int \sin^4(x) \cos(x) \, dx = \frac{\sin^5(x)}{5} + C \)

Partial Fraction Integration

The method called “partial fractions” is useful when the integrand contains a rational function, where a rational function is a fraction with polynomials in the numerator and denominator. Then the integrand can be rewritten as the sum of fractions and then each fraction is integrated separately.

The idea is to rewrite the rational function as a sum of simpler fractions, called partial fractions, which are easier to integrate individually. This method is especially useful when the denominator can be factored into linear or quadratic factors.

The process begins by factoring the denominator completely. Each factor is then assigned a corresponding term in the decomposition. For example, a linear factor like (x − a) gets a term A/(x − a), while a quadratic factor like (x² + bx + c) gets a term (Bx + C)/(x² + bx + c). After setting up the decomposition, coefficients are determined by clearing denominators and comparing coefficients of like terms, or by substituting convenient values for x. Once the decomposition is found, the integral is computed by integrating each simpler fraction separately.

Typically the following integration rule is used during the process:

\( \int \frac{1}{u} \, du = \ln|u| + C \)

Example:
Evaluate the integral:

\( \int \frac{1}{x^2 – 1} \, dx \)

Factor the denominator: \( x^2 – 1 = (x – 1)(x + 1) \).

The partial fraction decomposition is:
\( \frac{1}{x^2 – 1} = \frac{1}{2} \left[ \frac{1}{x – 1} – \frac{1}{x + 1} \right] \)

Now integrate each term separately:

\( \int \frac{1}{x^2 – 1} \, dx = \frac{1}{2} \left[ \ln|x – 1| – \ln|x + 1| \right] + C \)