Chapter 7: Techniques of Integration

Chapter 7: Techniques of Integration MATH 206-01: Calculus II Department of Mathematics University of Louisville last corrected September 14, 2013 1 / 43 Chapter 7: Techniques of Integration

7.1. Integration by Parts 7.1. Integration by Parts After completing this section, students should be able to: use integration by parts to evaluate indefinite and definite integrals use integration by parts multiple times solve application problems involving integrals where integration by parts is needed. 2 / 43 Chapter 7: Techniques of Integration

7.1. Integration by Parts Recall the Product Rule: d dx [f(x)g(x)] = f(x)g (x) + f (x)g(x) Integration by Parts is a method of integration derived from the product rule. {f(x)g (x) + f (x)g(x) } dx = f(x)g(x) (ignoring C for now) f(x)g (x) dx + f(x)g (x) dx = f(x)g(x) f (x)g(x) dx = f(x)g(x) f (x)g(x) dx 3 / 43 Chapter 7: Techniques of Integration

7.1. Integration by Parts Letting u = f(x) and v = g(x), it follows that du = f (x) dx and dv = g (x) dx. So the formula for integration by parts can also be expressed as u dv = uv v du. 4 / 43 Chapter 7: Techniques of Integration

7.1. Integration by Parts Example 7.1.1: Evaluate the following integrals. (a) xe x dx (b) arcsin t dt (c) r 3 sin 2r dr (d) e 3y cos y dy (e) w 3 e w2 dw Answers to Example 7.1.1: (a) (x 1)e x + C (b) t arcsin t + 1 t 2 + C (c) (r 3 /2 + 3r/4) cos 2r + (3r 2 /4 3/8) sin 2r + C (d) (sin y + 3 cos y)e 3y /10 + C (e) (w 2 + 1)e w2 /2 + C 5 / 43 Chapter 7: Techniques of Integration

7.1. Integration by Parts Integration by parts can also be applied to definite integrals: b a b a b f(x)g (x) dx = [f(x)g(x)] b a f (x)g(x) dx f(x)g (x) dx = f(b)g(b) f(a)g(a) Example 7.1.2: Evaluate 20 4 ln x dx. a b a f (x)g(x) dx Answer to Example 7.1.2: 20 ln 20 4 ln 4 16 38.369 6 / 43 Chapter 7: Techniques of Integration

7.1. Integration by Parts Example 7.1.3: Find the volume of the solid obtained by rotating the region bounded by y = sin(πx) for 0 x 1 2, x = 1 2, and y = 0 about the y-axis. Answer to Example 7.1.3: 2/π 7 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals 7.2. Trigonometric Integrals After completing this section, students should be able to: integrate trignometric integrals with products of powers of sine and cosine functions integrate trignometric integrals with products of powers of secant and tangent functions integrate trignometric integrals with products of sine and cosine functions with different angles solve application problems involving trigonometric integrals. 8 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals In this section, methods for integrating particular products of trigonometric functions are considered. In particular, we consider how to evaluate integrals of the forms sin m x cos n x dx tan m x sec n x dx for positive integers m and n. 9 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals Strategy for evaluating sin m x cos n x dx when the power of cosine is odd (n = 2k + 1, k 0): Save one cosine factor and express the factors in terms of sine by using the identity cos 2 x = 1 sin 2 x: sin m x cos 2k+1 x dx = sin m x (cos 2 x) k cos x dx = sin m x (1 sin 2 x) k cos x dx Then substitute u = sin x so that du = cos x dx to obtain u m (1 u 2 ) k du This also works if m is any real number. 10 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals Strategy for evaluating sin m x cos n x dx when the power of sine is odd (m = 2h + 1, h 0): Save one sine factor and express the factors in terms of cosine by using the identity sin 2 x = 1 cos 2 x: sin 2h+1 x cos n x dx = (sin 2 x) h cos n x sin x dx = (1 cos 2 x) h cos n x sin x dx Then substitute u = cos x so that du = sin x dx to obtain (1 u 2 ) h u n du This also works if n is any real number. 11 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals Strategy for evaluating sin m x cos n x dx when the powers of sine and cosine are both even (m = 2h and n = 2k, h > 0, k > 0): Use the half-angle identities sin 2 x = 1 2 (1 cos 2x) and cos 2 x = 1 2 (1 + cos 2x) to rewrite the integral as sin 2h x cos 2k x dx = (sin 2 x) h (cos 2 x) k dx = = { 1 2 1 2 h+k } h { 1 (1 cos 2x) (1 + cos 2x)) 2 (1 cos 2x) h (1 + cos 2x) k dx } k dx This also works either if m = 0 or if n = 0. 12 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals Example 7.2.1: Evaluate the following integrals. (a) (b) π/2 0 sin 2 θ cos 5 θ dθ cos 2 x sin 2 x dx Answers to Example 7.2.1: (a) 8/105 (b) x/8 sin(4x)/32 + C 13 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals Strategy for evaluating tan m x sec n x dx when the power of secant is even (n = 2k, k > 0): Save a factor sec 2 x and express the remaining factors in terms of tangent by using the identity sec 2 x = tan 2 x + 1: tan m x sec 2k x dx = tan m x (sec 2 x) k 1 sec 2 x dx = tan m x (tan 2 x + 1) k 1 sec 2 x dx Then substitute u = tan x so that du = sec 2 x dx to obtain u m (u 2 + 1) k 1 du This also works if m is any real number. 14 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals Strategy for evaluating tan m x sec n x dx when the power of tangent is odd (m = 2h + 1, h 0) and : Save a factor sec x tan x and express the remaining factors in terms of tangent by using the identity tan 2 x = sec 2 x 1: tan 2h+1 x sec n x dx = (tan 2 x) h sec n 1 x sec x tan x dx = (sec 2 x 1) h sec n 1 x sec x tan x dx Then substitute u = sec x so that du = sec x tan x dx to obtain (u 2 1) h u n 1 du This also works if n is any real number. 15 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals Example 7.2.2: Evaluate the following integrals. (a) (b) (c) π/4 0 π/3 0 sec 4 θ 3 tan θ dθ tan 3 t dt sec x dx Answers to Example 7.2.2: (a) 21/20 (b) 3/2 ln 2 (c) ln sec x + tan x + C 16 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals Strategy for evaluating integrals with products of sine and cosine with different angles: To evaluate the integral sin(mx) cos(nx) dx, use the identity sin A cos B = 1 [sin(a B) + sin(a + B)]. 2 To evaluate the integral sin(mx) sin(nx) dx, use the identity sin A sin B = 1 [cos(a B) cos(a + B)]. 2 To evaluate the integral cos(mx) cos(nx) dx, use the identity cos A cos B = 1 [cos(a B) + cos(a + B)]. 2 Odd/even functions: sin( θ) = sin θ and cos( θ) = cos θ 17 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals Example 7.2.3: Evaluate the following integrals. (a) sin(4x) cos(5x) dx (b) π/6 0 sin(2x) sin x dθ Answers to Example 7.2.3: (a) 1/2(cos(x) cos(9x)/9) + C (b) 1/12 18 / 43 Chapter 7: Techniques of Integration

7.3. Trigonometric Substitutions 7.3. Trigonometric Substitutions After completing this section, students should be able to: use trigonometric substitutions for integrating functions involving a 2 x 2, a 2 + x 2, or x 2 a 2 where a is a positive constant solve application problems involving trigonometric substitutions. 19 / 43 Chapter 7: Techniques of Integration

7.3. Trigonometric Substitutions Assume a > 0 so that a 2 = a = a. For problems involving a 2 x 2 : Let x = a sin θ for π 2 θ π 2. Then dx = a cos θ dθ and a 2 x 2 = a 2 (a sin θ) 2 = a 2 a 2 sin 2 θ = a 2 (1 sin 2 θ) = a 2 1 sin 2 θ = a 2 cos ( 2 θ = a cos θ since cos θ 0 for π 2 θ π ). 2 20 / 43 Chapter 7: Techniques of Integration

7.3. Trigonometric Substitutions Example 7.3.1: Evaluate the indefinite integral 9 x 2 dx x 2 Answer to Example 7.3.1: 9 x 2 /x arcsin(x/3) + C 21 / 43 Chapter 7: Techniques of Integration

7.3. Trigonometric Substitutions For problems involving a 2 + x 2 : Let x = a tan θ for π 2 < θ < π 2. Then dx = a sec 2 θ dθ and a 2 + x 2 = a 2 + (a tan θ) 2 = a 2 + a 2 tan 2 θ = a 2 (1 + tan 2 θ) = a 2 sec ( 2 θ = a sec θ since sec θ > 0 for π 2 < θ < π ). 2 22 / 43 Chapter 7: Techniques of Integration

7.3. Trigonometric Substitutions Example 7.3.2: Evaluate the definite integral 4 0 dx 9 + x 2 Answer to Example 7.3.2: ln 3 23 / 43 Chapter 7: Techniques of Integration

7.3. Trigonometric Substitutions For problems involving x 2 a 2 : Let x = a sec θ for 0 θ < π 2 or π θ < 3π 2. Then dx = a sec θ tan θ dθ and x 2 a 2 = (a sec θ) 2 a 2 = a 2 sec 2 θ a 2 = a 2 (sec 2 θ 1) = a 2 tan ( 2 θ = a tan θ since tan θ > 0 for 0 θ < π 2 or π θ < 3π ) 2 24 / 43 Chapter 7: Techniques of Integration

7.3. Trigonometric Substitutions Example 7.3.3: Evaluate the following integrals. (a) (b) (x 2 3) 3/2 dx x y 2 + 4y dy y + 2 Answers to Example 7.3.3: (a) (x 2 3) 3/2 /3 3 x 2 3 + 3 3 arcsec(x/ 3) + C (b) y 2 + 4y 2 arctan( y 2 + 4y/2) + C 25 / 43 Chapter 7: Techniques of Integration

7.3. Trigonometric Substitutions Example 7.3.4: Find the area of the region bounded above by y = 1, bounded below by y = 1 x 2, and bounded on the sides by x = 0 and x = 1. Answer to Example 7.3.4: 1 π/4 26 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions 7.4. Integration of Rational Functions by Partial Fractions After completing this section, students should be able to: use long division to express a rational function as a sum of a polynomial and a proper rational function use the method of partial fractions to rewrite a proper rational function integrate rational functions using the method of partial fractions and long division when needed solve application problems involving integration by partial fractions. 27 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions In this section, techniques are discussed which often help in the integration rational functions f(x) = P (x) Q(x) (where P and Q are polynomials). A rational function is said to be proper if the degree of P is less than the degree of Q. If a rational function is not proper, then long division can always be used to express it as a proper function of the form f(x) = P (x) R(x) = S(x) + Q(x) Q(x) where S and R are polynomials and the degree of R is less than the degree of Q. 28 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions Example 7.4.1: Evaluate the integral x 2 x 1 dx Answer to Example 7.4.1: x 2 /2 + x + ln(x 1) + C 29 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions The method of partial fractions is a very useful technique involved in the integration of proper rational functions. The Fundamental Theorem of Algebra implies that any polynomial can be expressed as a product of linear and quadratic terms. 30 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions Case 1: Suppose Q(x) can be factored as a product of distinct linear factors: Q(x) = (a 1 x + b 1 )(a 2 x + b 2 ) (a k x + b k ). Then partial fractions guarantees that there are constants A 1, A 2,..., A k such that R(x) Q(x) = A 1 + A 2 A k +... +. a 1 x + b 1 a 2 x + b 2 a k x + b k Example 7.4.2: Find the average value of the function f(x) = 1 4 x 2 on the interval [ 1, 1]. Answer to Example 7.4.2: ln 3/4 31 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions Case 2: Suppose Q(x) can be factored as a product of linear factors, some of which are repeated. Suppose a particular factor ax + b is repeated r times. Then the partial fractions decomposition will include the terms A 1 ax + b + A 2 (ax + b) 2 +... +... + A r (ax + b) r. Example 7.4.3: Evaluate the integral x 3 + 1 x 3 x 2 dx Answer to Example 7.4.3: x + 2 ln x 1 ln x + 1/x + C 32 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions Case 3: Suppose Q(x) includes irreducible quadratic factors, none of which are repeated. Suppose a particular factor ax 2 + bx + c is included. Then the partial fractions decomposition will include the term Ax + B ax 2 + bx + c. Example 7.4.4: Evaluate the integral 1 x 3 + x dx Answer to Example 7.4.4: ln x + ln(x 2 + 1)/2 + C 33 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions Case 4: Suppose Q(x) includes irreducible quadratic factors, some of which are repeated. Suppose a particular factor ax 2 + bx + c is repeated r times. Then the partial fractions decomposition will include the terms A 1 x + B 1 ax 2 + bx + c + A 2x + B 2 (ax 2 + bx + c) 2 +... +... + A rx + B r (ax 2 + bx + c) r. Example 7.4.5: Evaluate the integral 2x 2 3x + 2 x 4 + 2x 2 + 1 dx Answer to Example 7.4.5: 2 arctan x + 3/{2(x 2 + 1)} + C 34 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions Sometimes transformations can be made to general integrals that turn it into an integral of a rational function. Example 7.4.6: Evaluate the integral x + 4 dx x using the transformation u = x + 4. Answer to Example 7.4.6: 2 x + 4 + 2 ln x + 4 2 2 ln( x + 4 + 2) + C 35 / 43 Chapter 7: Techniques of Integration

7.8. Improper Integrals 7.8. Improper Integrals After completing this section, students should be able to: rewrite an improper integral as a limit or limits, determine if it is convergent, and evaluate it if it is convergent rewrite an integral with a discontinuous integrand as a limit or limits, determine if it is convergent, and evaluate it if it is convergent 36 / 43 Chapter 7: Techniques of Integration

7.8. Improper Integrals So far, when considering definite integrals b a f(x) dx, we have assumed that the interval [a, b] is finite and the function f is continuous on [a, b]. In this section, we extend the definition of the definite integral to cover cases when these assumptions are violated. 37 / 43 Chapter 7: Techniques of Integration

7.8. Improper Integrals Infinite Integrals: See illustrations on page 519. Suppose f is a continuous function on the interval [1, ) and we want to find the define Let A(t) = t 1 x = 1 to x = t. Then 1 1 f(x) dx. f(x) dx be the area under the curve from f(x) dx = A( ) = lim t A(t) = lim t t 1 f(x) dx. 38 / 43 Chapter 7: Techniques of Integration

7.8. Improper Integrals There are three types of definitions for infinite integrals: a b f(x) dx = lim f(x) dx = f(x) dx = t t a b lim t t c for any real number c f(x) dx f(x) dx f(x) dx + c f(x) dx An improper integral is called convergent if the corresponding limit exists. It is called divergent if the limit does not exist. 39 / 43 Chapter 7: Techniques of Integration

7.8. Improper Integrals Example 7.8.1: Determine whether each integral is convergent or divergent. Evaluate the integrals that are convergent. (a) (b) (c) 1 0 1 x dx xe 2x dx 1 4 + x 2 dx Answer to Example 7.8.1: (a) Divergent ( ) (b) 1/4 (c) π/2 40 / 43 Chapter 7: Techniques of Integration

7.8. Improper Integrals Discontinuous Integrands: Suppose f is a continuous function on the interval (0, 1] but discontinuous at 0, and we want to find the define 1 0 f(x) dx. Let A(t) = 1 x = t to x = 1. Then 1 0 t f(x) dx be the area under the curve from f(x) dx = A(0 + ) = lim A(t) = lim t 0+ t 0 + 1 t f(x) dx. 41 / 43 Chapter 7: Techniques of Integration

7.8. Improper Integrals There are three types of definitions for integrals with discontinuous integrands: If f is continuous on (a, b] but discontinuous at a, then b a f(x) dx = lim t a + b t f(x) dx. If f is continuous on [a, b) but discontinuous at b, then b a f(x) dx = lim t b t a f(x) dx. If f is discontinuous at c and c is in the interval (a, b), then b a f(x) dx = c a f(x) dx + b c f(x) dx. 42 / 43 Chapter 7: Techniques of Integration

7.8. Improper Integrals Example 7.8.2: Determine whether each integral is convergent or divergent. Evaluate the integrals that are convergent. (a) (b) (c) 1 0 3 0 1 1 ln x dx x 9 x 2 dx 1 x 3 dx Answer to Example 7.8.2: (a) 1 (b) 3 (c) Divergent (does not exist) 43 / 43 Chapter 7: Techniques of Integration