Calculus II MAT 146 Additional Methods of Integration

The methods of substitution and integration by parts are widely used methods of integration. Each of these methods is associated with a derivative rule. Substitution relies on undoing the chain rule and integration by parts results from undoing the product rule. Additional methods of integration are associated with particular types of functions. Here, we explore how to integrate certain trigonometric functions as well as rational functions.

Trigonometric Integrals: Using Trig Identities Shown below are four examples to illustrate integration of certain families of trigonometric integrals. 3 Example 1: Evaluate тsin xdx . Within these examples, sin 3 xdx notice the use of one or т more of the following 2 = т(sinx)(sin x)dx trig identities: 2 = т(sinx)(1 - cos x)dx Pythagorean Identities 2 cos2 x + sin2 x = 1 = т(sinx - sinx Чcos x)dx 2 2 2 1 + tan x = sec x = тsinxdx - т(sinx)(cos x)dx Half-Angle Identities The first integral in the last line can be solved by 1 inspection and the second by using the cos2 x = (1 + cos(2x)) substitution u = cosx. With this in mind, the final 2 result is 2 1 1 sin x = (1 - cos(2x)) sin 3 xdx = - cosx + cos3 x + C 2 т 3 2 Example 2: Evaluate тcos xdx . 2 тcos xdx 1 = (1 + cos(2x))dx т2 1 = (1 + cos(2x))dx 2 т 1 1 = dx + cos(2x)dx 2 т 2 т 1 1 ж1 ц = x + з sin(2x)ч + C 2 2 и2 ш 1 1 = x + sin(2x) + C 2 4

5 Example 3: Evaluate тsin xdx . 5 тsin xdx 2 2 = т(sinx)(sin x)(sin x)dx 2 2 = т(sinx)(1 - cos x)(1 - cos x)dx 2 4 = т(sinx)(1 - 2 cos x + cos x)dx 2 4 = т(sinx - 2sinx cos x + sinxcos x)dx 2 4 = тsinxdx - 2тsinxcos xdx + тsinxcos xdx The left integral in the last line can be solved by inspection, while each of the other two require a u-substitution of u = cosx. This gives us

5 2 4 тsin xdx = тsinxdx - 2тsinx cos xdx + тsinxcos xdx 2 1 = - cosx + cos3 x - cos5 x + C 3 5 4 Example 4: Evaluate тsin xdx . 4 тsin xdx 2 2 = тsin x Чsin xdx 1 2 = т[ 2 (1- cos(2x))] dx

1 2 = (1 - 2 cos(2x) + cos(2x))dx т4 1 1 1 = dx - cos(2x)dx + cos2 (2x)dx 4 т 2 т 4 т 1 1 1 1 = dx - cos(2x)dx + (1 + cos(4x))dx 4 т 2 т 4 т2 1 1 1 = dx - cos(2x)dx + (1 + cos(4x))dx 4 т 2 т 8 т 1 1 1 1 = dx - cos(2x)dx + dx + cos(4x)dx 4 т 2 т 8 т 8 т Each of the four integrals in the last line can be solved by inspection or by a straightforward substitution. This gives us 1 1 1 1 sin 4 xdx = dx - cos(2x)dx + dx + cos(4x)dx т 4 т 2 т 8 т 8 т 1 1 1 1 = x - sin(2x) + x + sin(4 x) + C 4 4 8 32 3 1 1 = x - sin(2x) + sin(4 x) + C 8 4 32

As we look back on these four examples of integrals involving powers of trigonometric functions, we can make some useful observations about strategies for evaluating such integrals. With integrals of the form n n тsin xdx or тcos xdx , where n is a positive integer greater than 1, we follow one of two strategies depending on whether the exponent is odd or even.

Situation A: If n = 2k + 1 for k some positive integer (i.e., n is an odd exponent) [see Examples 1 and 3 above]: 2k +1 2k 1. Factor the integrand from тsin xdx to т(sin x)sin xdx. 2 2 2k 2. Using the trig identity sin x + cos x = 1 , rewrite т(sin x)sin xdx as 2 k т(sin x)(1 - cos x) dx . 2 k 3. Expand (1- cos x) to get 2 k-1 2k-2 k 2k 1 - kcos x + L + (-1) kcos x + (-1) cos x . 4. Multiply each term in the above expansion by sinx to get the integral 2 k-1 2k-2 k 2k тsin x - ksinx cos x +L + (-1) ksinxcos x + (-1) sinx cos xdx 5. Integrate this expression term by term: 2 k-1 2k-2 k 2k т sin xdx - тksinxcos xdx +L + (-1) тksinx cos x + (-1) тsinxcos xdx 6. Evaluate the first term by inspection and make a substitution in all other terms, using u = cosx. 7. Evaluate the remaining integral expressions.

Situation B: If n = 2k for k some positive integer (i.e., n is an even exponent) [see Examples 2 and 4 above]: 2 1 1. Using the trig identity cos x = (1 + cos(2x)) , rewrite cos2k xdx as 2 т k й1 щ (1 + cos(2x)) dx , тлк2 ыъ k й1 щ 2. Expand (1 + cos(2x)) to get лк2 ыъ k-1 k 1 + kcos(2x) +L + kcos (2 x) + cos (2x) , 3. Integrate the previous expression term by term to get k k ж1 ц ж1 ц з ч dx + з ч Чk cos(2x)dx +L и2 ш т и2 ш т k k ж1 ц k-1 ж1 ц k +з ч Чkтcos (2x)dx + з ч тcos(2x)dx и2 ш и2 ш 4. For each term in the previous sum, either (a) evaluate by inspection, (b) return to step (1) here and use the same trig identity (even exponents), or (c) return to Situation (A) previously described and follow those steps (odd exponents). Continue this process until each integral created can be solved by inspection. Rational-Function Integrals: Using Partial Fraction Decomposition

Which of the following integrals is easier to evaluate? 1 1 1 (A) dx or (B) 1 dx - 1 dx тx 2 - 5x - 6 7 тx - 6 7 тx + 1 For most of us, it is (B). Each of the two integrals in (B) is of the form du C , where C is some constant and u is a function of x. The integral in (A) т u does not have a straightforward strategy for solution unless the denominator matches a special case.

It turns out, however, that (A) and (B) are equivalent, because 1 1 1 2 = - x - 5x - 6 7(x - 6) 7(x +1) You can verify this by adding the two fractions in the right-side expression. After simplifying the sum, it will be the left-side expression.

In order to evaluate integrals such as those illustrated in (A) above, it is useful to be able to transform a rational expression into a sum of difference of simpler rational expressions. This transformation process, essentially reversing the process of determining a common denominator and then adding or subtracting, is called partial fraction decomposition. Here we describe how to carry out partial fraction decomposition and then use that process to transform integrals of rational expressions into integrals that are equivalent yet more easily evaluated.

The format for a partial fraction decomposition depends on the characteristics of the denominator in a rational expression. In the expression 1 , we can factor the denominator into two linear terms: x2 - 5x - 6 1 1 = x2 - 5x - 6 (x - 6)(x + 1) If a rational expression contains two linear terms in its denominator—linear terms that are not multiples of one another—the expression can be transformed into an equivalent expression as follows: 1 A B = + (x - 6)(x + 1) (x - 6) (x +1) If we multiply each side of this equation by (x-6)(x+1), we get: 1 = A(x +1) + B(x - 6) 0x + 1 = Ax+ A+ Bx - 6B 0x + 1 = (A+ B)x + (A- 6B) Note that we have rewritten the left side expression as 0x+1. We do this in order to equate the linear and constant coefficients on the two sides of the equal sign: 0x + 1 = (A+ B)x + (A- 6B) Ю A+ B= 0 Ю A- 6B= 1 The last two equations for a system that we can solve using techniques learned in algebra: A - 6B = 1Ю A = 1 + 6B 1 A+ B = 0 (1 + 6B) + B = 0 7B= -1 B = - Ю Ю Ю 7 ж 1ц 1 Ю A+ з - ч = 0 Ю A = и 7ш 7 This shows us that 1 1 - 1 1 1 = 7 + 7 = - (x - 6)(x + 1) (x - 6) (x +1) 7(x - 6) 7(x + 1)

Example 1: Use partial fraction decomposition to transform the integral 5x - 4 dx into a form more easily integrated and then carry out the т2x2 + x -1 integration. 5x - 4 5x - 4 A B = = + 2x2 + x - 1 (x + 1)(2x -1) (x + 1) (2x - 1) Ю 5x - 4 = A(2x - 1) + B(x + 1) Ю 5x - 4 = 2 Ax- A+ Bx + B Ю 5x - 4 = (2A + B)x + (B- A) Ю 2A+ B = 5 and B- A = -4 The final two equations form a linear system whose solution is A = 3 and 5x - 4 ж 3 1 ц B = -1. Therefore, dx = з - чdx . т2x2 + x -1 ти x + 1 2x -1ш We can evaluate the new integral by inspection or by using a separate u- substitution for each fraction. In so doing, the final result is ж 3 1 ц з - ч dx = 3lnx + 1 - 1 ln2x -1 + C . тиx +1 2x - 1ш 2

10x + 3 Example 2: Evaluate dx . т x2 + x 10x + 3 10x + 3 A B = = + x2 + x x(x + 1) x x +1 Ю 10x + 3 = A(x +1) + B(x) Ю 10x + 3 = Ax+ A+ Bx Ю 10x + 3 = (A+ B)x + (A) Ю A+ B= 10 and B = 3 Ю A = 7 The linear system has solution A = 7 and B = 3. Therefore, 10x + 3 ж7 3 ц dx = з + чdx . We can evaluate the new integral by т x2 + x тиx x +1ш ж7 3 ц inspection. The final result is з + чdx = 7 lnx + 3 lnx + 1 + C . тиx x + 1ш

1 x 3 Example 3: Evaluate т 2 dx . 0 x +1 Because the numerator is a higher-power polynomial than the denominator, we first carry out long division on the rational expression. This gives us x 3 x = x - . We now explore how to solve the new integrals: x2 + 1 x2 +1 1 ж x ц 1 1 x тз x - 2 чdx = тxdx -т 2 dx . 0 и x +1ш 0 0 x +1 We can evaluate the first integral by inspection. In the second, letting 2 1 u = x +1 with du = 2xdx or /2du = xdx leads to an integral we can solve by inspection. We get 1 1 1 x x2 1 xdx - dx = - lnx2 + 1 т тx2 + 1 2 2 0 0 0 ж1 1 ц ж 1 ц = з - ln2ч - з 0 - ln1ч и2 2 ш и 2 ш 1 = (1 - ln2) 2 Note 1: On pages 406-407 in our textbook, the author describes various formats and requirements for use of partial fraction decomposition that depend on the characteristics of the denominator in the initial rational expression.

Note 2: The expand function on your TI-89 or TI-92 calculator will perform a partial fraction decomposition before your eyes! Type expand( and then type in the rational expression and close the parentheses. When you hit return, the decomposition will be shown on your screen. Among other locations, the expand( command can be found under the algebra menu. Touch the F2 key and then select choice 3.