Chapter 8 8.2 Integration By Parts Let G(x) be any antiderivative of g(x). In this case G’(x)=g(x), then by the product rule, Which implies Or, equivalently, as (1) The application of this formula is called integration by parts.

In practice, we usually rewrite (1) by letting This yields the following alternative form for (1)

Example: Use integration by parts to evaluate Solution: Let

Guidelines for Integration by Parts The main goal in integration by parts is to choose u and dv to obtain a new integral That is easier to evaluate than the original. A strategy that often works is to choose u and dv so that u becomes “simpler” when Differentiated, while leaving a dv that can be readily integrated to obtain v. There is another useful strategy for choosing u and dv that can be applied when the Integrand is a product of two functions from different categories in the list. Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential In this case, you will often be successful if you take u to be the function whose Category occurs earlier in the list and take dv to the rest of the integrand (LIATE). This method does not work all the time, but it works often enough to b e useful.

Example. Evaluate Solution: According to LIATE, we should let

Example: Evaluate Solution: Let

Repeated Integration by Parts Example: Evaluate Solution: Let Apply integration by parts to

Finally,

Example: Evaluate Solution: Let (a)

Together with (a), we have Solve for the unknown integral, we have

Integration by Parts for Definite Integrals For definite integrals, the formula corresponding to is

Example: Evaluate Solution: Let

8.3 Trigonometric Integrals Integrating powers of sine and cosine By applying the integration by parts , we have two reduction formulas

In particular, ……

Integrating Products of Sines and Cosines

Example: Evaluate Solution: since n=5 is odd,

Integrating Powers of Tangent and Secant There are similar reduction formulas to integrate powers of tangent and secant.

In particular, ……

Integrating Products of Tangents and Secents (optional)

Section 8.4 Trigonometric Substitutions We are concerned with integrals that contain expressions of the form In which a is a positive constant. The idea is to make a substitution for x that will eliminate the radical. For example, to eliminate the radical in , we can make the substitution x = a sinx, - /2    /2 Which yields

Example Example 1. Evaluate Solution. To eliminate the radical we make the substitution x= 2sin, dx= 2cosd This yields

Example Cont. From the left figure, we obtain Substituting this in the previous answer yields

Example There are two methods to evaluate the definite integral. Make the substitution in the indefinite integral and then evaluate the definite integral using the x-limits of integration. Make the substitution in the definite integral and convert the x-limits to the corresponding -limits. Example. Evaluate Solution: Since the substitution can be expressed as =sin-1(x/2), the -limits of the integration are x=1: = sin-1(1/2)=/6 x= : = sin-1( /2)=/4 Thus

Method of Trigonometric Substitution

Example dx= 5sec tan d. Thus, Example. Evaluate , assuming that x5. Solution. Let x=5 sec, 0  </2 dx= 5sec tan d. Thus, Note that

8.5 Integrating Rational Functions by Partial Fractionas Suppose that P(x) / Q(x) is a proper rational function, by which we mean that the Degreee of the numerator is less than the degree of the denominator. There is a Theorem in advanced algebra which states that every proper rational function can Be expressed as a sum Where F1(x), F2(x), …, Fn(x) are rational functions of the form In which the denominators are factors of Q(x). The sum is called the partial fraction Decomposition of P(x)/Q(x), and the terms are called partial fraction.

Finding the Form of a Partial Fraction Decomposition The first step in finding the form of the partial fraction decomposition of a proper Rational function P(x)/Q(x) is to factor Q(x) completely into linear and irreducible Quadratic factors, and then collect all repeated factors so that Q(x) is expressed as A product of distinct factors of the form From these factors we can determine the form of the partial fraction decomposition Using two rules that we will now discuss.

Linear Factors If all of the factors of Q(x) are linear, then the partial fraction decomposition of P(x)/Q(x) can be determined by using the following rule:

Example: Evaluate Solution: By the linear factor rule, the decomposition has the form Where A and B are constants to be determined. Multiplying this expression Through by (x-1)(x+2) yields 1 = A ( x + 2 ) + B ( x – 1 ) Let x=1, we have 1= 3A which implies A = 1/3; and let x = -2, we have 1 = -3B, Which means B = -1/3. Thus,

The integration can now be completed as follows:

Example: Evaluate Solution: By the linear factor rule, Multiplying by yields (1) (2) Setting x=0 in (1) yields B = -2, and setting x = 2 in (1) yields C = 2. Then By equating the coefficient of x2 on the two sides to obtain A + C = 0 or A = -C = -2

Substituting A = -2, B = -2, and C = 2 yields the partial fraction decomposition Thus

Quadratic Factors If some of the factors of Q (x) are irreducible quadratics, then the contribution of Those facctors to the partial fraction decomposition of P(x)/Q(x) can be determined From the following rule: