Math02 - Term72 - Guides and Exercises - DRAFT 7 Techniques of Integration A summery for the most important integrals that we have learned so far: 7. Integration by Parts The Product Rule states that if f and g are differentiable functions, then (f(xg(x = f(x g(x + f(xg(x and integrating both sides, we get f(xg(x = (f (xg(xdx + (f(xg (xdx. Arranging this equation, we get the integration by part method f(xg(x (f (xg(xdx = (f(xg (xdx. Let u = f(x and v = g(x. Then the differentials are du = f (xdx and dv = g (xdx. By the Substitution Rule, the formula for integration by parts becomes DRAFT - Math 02 Lecture Note - Note 6. The integral u dv can be obtained in terms of to evaluate than u dv. v du with the hope that v du is easier Page 48
Math02 - Term72 - Guides and Exercises - DRAFT Steps to use integration by part method i Factor the integrand into two parts, u and dv. Make your choice based on these guidelines: a the antiderivative v = dv is easy to find b the new integral v du is easier to evaluate than the original one. c Sometimes, dv = dx is a good choice. ii Apply the formula u dv = uv v du. Group Example 7.. Find i sin x dx ii ln x dx iii tan x dx DRAFT - Math 02 Lecture Note - iv π/3 0 sin x ln(cos x dx Page 49
Math02 - Term72 - Guides and Exercises - DRAFT Group 2 Example 7.2. Find i x e x dx ii x sin x dx iii x 7 cos x dx (using the Tabular Group 3 Example 7.3. Find i e x sin(x dx DRAFT - Math 02 Lecture Note - ii π 0 e 3x sin(2x dx Page 50
Math02 - Term72 - Guides and Exercises - DRAFT Reduction Formula Example 7.4. Drive a reduction formula for i sec n x dx for n 3. ii sin n x dx for n 3. (same idea as i DRAFT - Math 02 Lecture Note - Page 5
Math02 - Term72 - Guides and Exercises - DRAFT Exercise and old Exam Questions x tan x dx = a 2 ( x 2 tan x + tan x x + C 2 3 b 2 ( x tan x x 2 + C c 2 x2 tan x tan x + C d x 2 tan x tan x + C x e x 2 + + tan x + C e θ cos(2θ dθ = a 5 e θ sin(2θ 5 e θ cos(2θ + c b 5 e θ sin(2θ 2 5 e θ cos(2θ + c c 2 5 e θ sin(2θ + 5 e θ cos(2θ + c d 2 5 e θ sin(2θ 5 e θ cos(2θ + c e 2 5 e θ sin(2θ 2 5 e θ cos(2θ + c π2 4 0 cos( x dx = a π 2 2 b π 2 + c 2 π DRAFT - Math 02 Lecture Note - d π 3 2 e π 2 4 0 x 3 e x dx = a 6 2e b 6 3e c 6 4e d 6 + 2e e 6 + 3e Page 52
Math02 - Term72 - Guides and Exercises - DRAFT 7.2 Trigonometric Integrals We use trigonometric identities to integrate certain combinations of trigonometric functions, like: sin 3 x cos 2 x dx tan 3 x sec 2 x dx sin(2x cos(3x dx Group I: Integrals of form sin m x cos n x dx where m or n is odd DRAFT - Math 02 Lecture Note - Example 7.5. Find i cos 5 x dx Page 53
Math02 - Term72 - Guides and Exercises - DRAFT ii sin 3 x cos x dx Group II: Integrals of form Example 7.6. Find i sin 2 x cos 4 x dx ii sin 2 (5x cos 2 (5x dx sin m x cos n x dx both m and n are even DRAFT - Math 02 Lecture Note - Page 54
Math02 - Term72 - Guides and Exercises - DRAFT Group III: Integrals of form tan m x sec n x dx where n is even Example 7.7. Find i tan 3/2 x sec 4 x dx ii cot 3/2 csc 4 x dx Group IV: Integrals of form tan m x sec n x dx where m is odd DRAFT - Math 02 Lecture Note - Example 7.8. Find i tan 5 x sec /2 x dx Page 55
Math02 - Term72 - Guides and Exercises - DRAFT ii cot 5 x csc /2 x dx Group V: Integrals of form sin(mx cos(nx dx, Example 7.9. Find i sin(8x cos(5x dx ii π/2 0 cos(5x cos(0x dx sin(mx sin(nx dx, or DRAFT - Math 02 Lecture Note - cos(mx cos(nx dx Note 7. You should be able to drive (justify these formulas: i tan x dx = ln sec x + C ii sec x dx = ln sec x + tan x + C Page 56
Math02 - Term72 - Guides and Exercises - DRAFT Exercise and old Exam Questions 4 tan 3 x dx = 2 a tan 4 x + c b 2 tan 2 x + ln cos x + c c 2 tan 2 x + 4 ln cos x + c d 2 tan 2 x + cot x cos 2 x + c π 2 0 e 4 tan 2 x + ln cos x + c sin 3/2 x cos 3 x dx = a 4 5 8 b 45 c 5 9 d 45 e 3 45 3 If 3 sec x tan 3 x dx = A sec α x + B sec β x + C, where α, β, A, B are real numbers, and C is an arbitrary constant, then the product αβa B is equal to a b 2 c 0 d 7 e 2 DRAFT - Math 02 Lecture Note - Page 57
Math02 - Term72 - Guides and Exercises - DRAFT 7.3 Trigonometric Substitution We need a special substitution to integrate functions involving one of the square root expressions: a 2 x 2, x 2 + a 2, or x 2 a 2. Steps to apply Trigonometric Substitution Step. Substitute to eliminate the square root. Step 2. Evaluate the trigonometric integral. Step 3. Convert back to original variable using the right triangle. Group I: Integrals involving a 2 x 2 Example 7.0. Find i x 2 dx DRAFT - Math 02 Lecture Note - ii (4 9x 2 dx 3/2 Page 58
Math02 - Term72 - Guides and Exercises - DRAFT Group II: Integrals involving x 2 + a 2 Example 7.. Find i x 2 x 2 + 4 dx ii 3x 3 (6x 2 + 9 3/2 dx Group III: Integrals involving x 2 a 2 DRAFT - Math 02 Lecture Note - Example 7.2. Find i x 3 x 2 dx Page 59
Math02 - Term72 - Guides and Exercises - DRAFT ii 2 2 x 3 x 2 dx Note 8. Integrands involving x 2 + bx + c are treated by completing the square. Example 7.3. Find i (x 2 6x + 2 dx ii dx 5 4x x 2 DRAFT - Math 02 Lecture Note - Page 60
Math02 - Term72 - Guides and Exercises - DRAFT Exercise and old Exam Questions x x 4 dx = a x + x 2 x 4 + c 2 b 4 (x2 x 4 + sin (x 2 + c c 4 (x2 x 4 3 sin (x 2 + c d 2 x 4 + 2 sin (x 2 + c e x 4 + sin (x 2 + c 36 x 2 dx = a 9 ( 36 x 2 sin x + C ( 6 b 8 sin x + 36 x 6 2 + C ( c 8x sin x + C ( 6 d 9 sin x 2x 36 x 6 2 + C ( e 8 sin x + 6 2 x 36 x 2 + C x 3 If x > 3, then 2 9 x 3 dx = a ( x x 6 sec + 2 9 + c 3 2x b ( x x 3 sec 2 9 3 2x 2 + c c ( x x 3 sec + 2 9 3 x 2 + c d ( x x 6 sec 2 9 3 2x 2 + c e ( x x 6 sec + 2 9 + c 2 2x DRAFT - Math 02 Lecture Note - Page 6
Math02 - Term72 - Guides and Exercises - DRAFT 7.4 Integration of Rational Functions by Partial Fractions We introduce the Method of Partial Fractions, used to integrate rational quotient of two polynomials functions P (x P (x Q(x, where P (x and Q(x are polynomials. A rational function Q(x is called proper if the degree of P (x less than the degree of Q(x. For example If P (x Q(x is not proper (improper, use long division i.e. P (x r(x Q(x = s(x + Q(x also polynomials and now r(x Q(x is proper. If P (x Q(x is proper, make use of the partial fraction method (coming. where s and r are Now, assume that we have proper rational functions deg.(p (x deg.(q(x. We consider the following cases: Case I: The denominator Q(x is a product of distinct (no repeated linear factors. It means that the denominator Q(x can be factored out as Q(x = (a x+b (a 2 x+b 2... (a k x+b k. In this case the partial fraction theorem states that there exist constants A, A 2,..., A k such that P (x Q(x = A + A 2 A k + +. a x + b a 2 x + b 2 a k x + b k It remains to determine the values of these constants. Example 7.4. Find x 3 + i x 2 4 dx DRAFT - Math 02 Lecture Note - ii x 2 + 2x 2x 3 + 3x 2 2x dx Page 62
Math02 - Term72 - Guides and Exercises - DRAFT Case II: Q(x is a product of linear factors, some of which are repeated. Suppose the first linear factor (a x + b is repeated r times; that is, (a x + b r occurs in the factorization of Q(x. Then instead of the single term A A a x+b we would include a x+b + A 2 (a x+b + 2 A... r (a x+b. For example, = A r x 3 (x 5 2 x + A 2 + A x 2 3 + B x 3 x 5 + B 2. (x 5 2 Example 7.5. Find 3x 9 i (x (x + 2 2 dx ii x 4 2x 2 + 4x + x 3 x 2 x + dx Case III: Q(x contains irreducible quadratic factors, none of which is repeated. Irreducible quadratic expression like x 2 + and x 2 + x + 4. If Q(x has the factor ax 2 + bx + c, where b 2 4ac < 0, then, in addition to the known partial fractions, the expression will have a term of the form Ax+B where A and B are constants to be ax 2 +bx+c determined. For example, DRAFT - Math 02 Lecture Note - Example 7.6. Find 8x i x(x + 3(x 2 + 9 dx x 3 (x 2 + = A x + A 2 + A x 2 3 + B x+b 2 x 3 x 2 +. Page 63
Math02 - Term72 - Guides and Exercises - DRAFT ii 2x 2 x + 4 x 3 + 4x dx iii x 2 + x(x 2 + 3 dx* Case IV: Q(x contains a repeated irreducible quadratic factor. If Q(x has the factor s (ax 2 + bx + c r, where b 2 4ac < 0, then instead of the single partial fraction, we will have A x + B ax 2 + bx + c + A 2 x + B 2 (ax 2 + bx + c 2 + + A r x + B r (ax 2 + bx + c r. Note 9. Each of the terms in the above expression can be integrated by using a substitution or by first completing the square if necessary. 4 x Example 7.7. Find x(x 2 + 2 2 dx DRAFT - Math 02 Lecture Note - Example 7.8. Write out the form of the partial fraction decomposition of x(x + x 3 (x 2 (x 2 + 3x + 2(x 2 + 4 2 do not determine the numerical values of the coefficients Page 64
Math02 - Term72 - Guides and Exercises - DRAFT Rationalizing Substitutions If an integrand contains an expression of the form n h(x, then the substitution u = n h(x will mostly simplify the integrand. Example 7.9. Find i ii 0 + 3 x dx x 3 x dx Special Substitution by Karl Weierstrass (85 897 Letting t = tan ( ( x 2, π < x < π, one can show that cos x 2 = also, cos x = t2 + t 2, and sin x = 2t + t 2, +t 2 and sin ( x DRAFT - Math 02 Lecture Note - dx = 2 dt + t 2. 2 = t +t 2 Example 7.20. Use the above substitution to transform the integrand into a rational function of t and then evaluate the integral i 3 sin x 4 cos x dx and Page 65
Math02 - Term72 - Guides and Exercises - DRAFT ii π/2 0 sin 2x 2 + cos x dx DRAFT - Math 02 Lecture Note - Page 66
Math02 - Term72 - Guides and Exercises - DRAFT Exercise and old Exam Questions csc 2 t cot 2 t + cot t dt = 2 3 4 a ln + cot t + C b ln cot 2 t + cot t + C c ln + tan t + C d cos t + sin(2t + C e sin t + cot t + C 3x 3 3x 2 + 4 x 2 dx = x a 3 2 x2 + 4 ln x x + c b 3 2 x2 + 8 ln x x + c c 3 2 x2 + 2 ln x 2 x + c d 3x 2 + 2 ln x x + c e 3x 2 + 2 ln x 2 x + c x 2 + x + x 3 dx = + x a ln x + ln(x 2 + 2 tan x + c b ln 2x ln(x 2 + + tan x + c c ln x + 2 ln(x 2 + + tan x + c d ln x ln(x 2 + + tan x + c e ln 2x + ln(x 2 + tan x + c 0 (x (x 2 + 9 dx = DRAFT - Math 02 Lecture Note - a ln x 2 ln(x2 + 9 ( x 3 tan + c 3 b ln x 3 ln(x2 + 9 + ( x 3 tan + c 3 c ln x + 2 ln(x2 + 9 + ( x 3 tan + c 3 d ln x + 3 ln(x2 + 9 ( x 3 tan + c 3 e ln x 3 ln(x2 + 9 ( x 3 tan + c 3 Page 67
Math02 - Term72 - Guides and Exercises - DRAFT 7.5 Strategy for Integration Basic Integration Formulas Strategy for Integration. Simplify the integrand if possible 2. Look for an obvious substitution 3. Classify the integrand according to its form a Trigonometric functions (7.2 b Rational functions: you need partial fractions. (7.4 DRAFT - Math 02 Lecture Note - c Integration by parts (7. d Radicals: you may use trigonometric substitution (7.3 or rationalization. 4. try again : Example 7.2. Find i tan 3 x cos 3 x dx Page 68
Math02 - Term72 - Guides and Exercises - DRAFT ii e x dx iii iv x ln x dx x + x dx Q: Can We Integrate All Continuous Functions? DRAFT - Math 02 Lecture Note - Page 69
Math02 - Term72 - Guides and Exercises - DRAFT Exercise and old Exam Questions 2 3 4 2 0 x 4 + x 4 dx = a 2 + 2 ln 2 b 2 + 4 ln 2 c + 2 ln 2 d + 2 ln 2 e 3 + 4 ln 2 x x 2 + dx = a ln x 2 + ln x + C b 2 x2 + ln( + x 2 + + C c ln + x 2 + x + C d x ln x 2 + + C e x 2 + + C x (x 2 + sech(ln x dx = a x 2 ln x + tanh(ln x + c b x 3 + c c x 2 + c d sech(ln x + x 2 sech(ln x + c ( x 3 e 3 + x sech(ln x + c sin (e x e 2x dx = DRAFT - Math 02 Lecture Note - a 2 e x (sin (e x 2 + c b 3 2 (sin (e x 2 + c c 2 (sin (e x 2 + c d e x sin (e x + c e (sin (e x 2 + c 7.6 Integration Using Tables and Computer Algebra Systems 7.7 Approximate Integration Page 70
Math02 - Term72 - Guides and Exercises - DRAFT 7.8 Improper Integrals Motivation Example Example 7.22. Find x 2 dx So far, we know that if f is a continuous on a bounded interval [a, b] then DRAFT - Math 02 Lecture Note - b a f(xdx exists. It is called proper integral. In this section, we introduce three types of improper integrals: Type The integrand is a continuous function but one (or both limit(s of integration is(are infinite such as: e x dx, e x dx, e x dx. Type 2 The integrand has an infinite discontinuity on the interval [a, b] such as: 4 (x 2 2 dx, 2 0 (x 2 2 dx. Type 3 A mixture of types I and II. 4 2 (x 2 2 dx, We will use limits to find weather an improper integral exists (called converge or does not exist (called diverge. Definition 7.. Improper integral of type I. Note 0. f(x dx = a f(x dx + a f(x dx if both integrals are convergent. Page 7
Math02 - Term72 - Guides and Exercises - DRAFT Example 7.23. Determine weather the following integrals convergent or divergent? i ii iii π/2 x e x2 dx cos x dx DRAFT - Math 02 Lecture Note - + x 2 dx Example 7.24. For what values of p is the integral dx convergent? Page 72
Math02 - Term72 - Guides and Exercises - DRAFT Definition 7.2. Improper integral of type II. Note. b a f(x dx = c a f(x dx + b c f(x dx if both integrals are convergent. Example 7.25. Determine weather the following integrals convergent or divergent? i ii 2 0 dx 4 x 2 x + dx DRAFT - Math 02 Lecture Note - iii 3 0 dx (x 2/3 Page 73
Math02 - Term72 - Guides and Exercises - DRAFT A Comparison Test for Improper Integrals Theorem 7.. Suppose that f and g are continuous functions with f(x g(x 0 for all x a. i If ii If b a b a f(x dx is convergent, then g(x dx is divergent, then b b a a g(x dx is convergent. f(x dx is divergent. Example 7.26. Determine weather the following integrals convergent or divergent? i ii π/2 cos 2 x dx x 3 5 + 3e x dx x DRAFT - Math 02 Lecture Note - Page 74
Math02 - Term72 - Guides and Exercises - DRAFT Exercise and old Exam Questions The improper integral 0 dx is 2 3x a divergent b convergent and its value is ln 2 c convergent and its value is 0 d convergent and its value is 3 e convergent and its value is ln 5 2 The improper integral a p < b p > 2 c p > d p < 0 e p > 0 3 The improper integral 3π/2 a convergent and its value is 0 b convergent and its value is ln 2 c convergent and its value is ln 2 d divergent e convergent and its value is 2 4 The improper integral a equal to 2 3 0 x( + x 2 2p+ dx is convergent if sin x dx is + cos x dx is (4 + 3x 3/2 DRAFT - Math 02 Lecture Note - b equal to 3 4 c equal to d divergent e equal to 4 3 Page 75