Chapter 7 Notes, Stewart 7e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m xcos n (x)dx...

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1 Contents 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m xcos n (x)dx Evaluating tan m xsec n xdx Evaluating tanxdx, cotxdx, secxdx, and cscxdx Trigonometric Substitution for Integrals Involving a 2 x 2, a 2 +x 2, x 2 a Integration of Rational Functions by Partial Fractions Reducing an Improper Fraction (long division) Partial Fractions Rationalizing substitutions Additional Examples Strategy for Integration Basic Integration Formulas Procedures for matching integrals to basic formulas Integral Tables Approximate Integration Midpoint Rule Trapezoidal Rule Simpson s / Parabolic Rule

2 7.1 Integration by Parts Introduction If the techniques/formulas we introduced earlier do not work then there is another technique that you can use: Integration by Parts. It is a way of simplifying integrals of the form f(x)g(x)dx in which f(x) can be differentiated repeatedly and g(x) can be integrated repeatedly without difficulty. The Formula A. Derivation of the Formula The formula for integration by parts comes from the Product Rule. d (uv) = udv dx dx +vdu dx d(uv) = udv +vdu (differential form) udv = d(uv) vdu (Solve for udv ) udv = uv v du (integrating both sides) B. Note: The integration by parts expresses one integral, vdu in terms of another, udv. The idea is that with a proper choice of u and v, the second integral is easier to integrate than the original. You may have to use this technique several times before you reach an integral that you can integrate easily. C. The Integration-by-Parts Formula 1. Indefinite Integrals If u(x) and v(x) are functions of x and have continuous derivatives, then udv = uv vdu 2. Definite Integrals D. How to pick u and dv b a udv = uv When deciding on your choice for u and dv, use the acronym: ILATE with the higher one having the priority for u. I -inverse trig functions L -logarithmic functions A -algebraic functions T -trigonometric functions E -exponential functions b a b a vdu 2

3 Example xe 2x dx Example lnxdx Example e 1 x 2 lnxdx 3

4 Example tan 1 xdx Example y 2 e 3y dy 4

5 Example t 4 sin(2t)dt Example e x cos(4x)dx Example Use the reduction formula sec 4 ydy sec n udu = 1 n 1 tanusecn 2 u+ n 2 n 1 sec n 2 udu 5

6 Example t 3 cos(t 2 )dt Example Find the area of the region enclosed by the curve y = xcosx and the x-axis from x = π 2 to x = 3π 2 y y = xcos(x) π 2 π 3π 2 x 3 6

7 Example Find the volume of the solid generated by revolving the region in the 1 st quadrant bounded by the axes, the curve y = e x and the line x = ln2 about the line x = ln2. 2 y = e x, x = ln2, x = 0, y = 0 1 ln2 1 x A Summary of Common Integrals Using Integration by Parts: x n e ax dx x n sin(ax)dx x n cos(ax)dx x n lnxdx x n sin 1 (ax)dx x n tan 1 (ax)dx e ax sin(bx)dx e ax cos(bx)dx 7

8 7.2 Trigonometric Integrals Evaluating sin m xcos n (x)dx A. If the power of cosine is odd (n = 2k+1), save one cosine factor and use cos 2 x = 1 sin 2 x to express the remaining factors in terms of sine: sin m xcos 2k+1 xdx = sin m xcos 2k xcosxdx = sin m x ( cos 2 x ) k cosxdx = sin m x ( 1 sin 2 x ) k cosxdx Then substitute u = sinx. B. If the power of sine is odd (m = 2k+1), save one sine factor and use sin 2 x = 1 cos 2 x to express the remaining factors in terms of cosine: sin 2k+1 xcos n xdx = sin 2k xcos n xsinxdx = = (sin 2 x ) k cos n xsinxdx (1 cos 2 x ) k cos n xsinxdx Then substitute u = cosx. C. If the powers of both sine and cosine are odd, either 1 or 2 can be used. D. If the powers of both sine and cosine are even, use the half-angle identities sin 2 x = 1 2 [1 cos(2x)] and cos2 x = 1 2 [1+cos(2x)] E. It is sometimes helpful to use the identity sinxcosx = 1 2 sin(2x). Example sinθcos 4 (θ)dθ 8

9 Example sin 3 φdφ Example sin 3 xcos 3 x x dx 9

10 Example Find the area bounded by the curve of y = sin 4 (3x) and the x-axis from x = 0 to x = π 3 y y = sin 4 (3x) 1 π 3 x Evaluating tan m xsec n xdx A. If the power of secant is even (n = 2k), save one factor of sec 2 x and use sec 2 x = 1+tan 2 x to express the remaining factors in terms of tangent: tan m xsec 2k xdx = = = tan m xsec 2k 2 xsec 2 xdx tan m x ( sec 2 x ) k 1 sec 2 xdx tan m x ( 1+tan 2 x ) k 1 sec 2 xdx Then substitute u = tanx. 10

11 B. If the power of tangent is odd (m = 2k +1), save one factor of secxtanx and use tan 2 x = sec 2 x 1 to express the remaining factors in terms of secant: tan 2k+1 xsec n xdx = tan 2k xsec n 1 x secxtanxdx = = (tan 2 x ) k sec n 1 x secxtanxdx (sec 2 x 1 ) k sec n 1 x secxtanxdx Then substitute u = secx. Example tan 6 ydy Example π/6 0 tanθsec 3 θdθ 11

12 Example Find the volume of the solid generated by revolving the region bounded by the curve of y = sec 2 (x 2 )tan 2 (x 2 ), the x-axis and the line x = π 2 about the y-axis. y y = sec 2 (x 2 )tan 2 (x 2 ) 2 1 x π Evaluating tanxdx, cotxdx, secxdx, and cscxdx 1. When evaluating use u-substitution. 2. When evaluating u-substitution. 3. When evaluating u-substitution. tanxdx or cotxdx rewrite the integrand in terms of sine and cosine and then secxdx multiply the integrand by the form of one, cscxdx multiply the integrand by the form of one, secx+tanx, and then use secx+tanx cscx+cotx, and then use cscx+cotx 12

13 Example π/2 π/4 cotθdθ Example secxdx Integral Formulas for tan(x), cot(x), sec(x), csc(x) 1. tanudu = ln cosu +C = ln secu +C cotudu = ln sinu +C = ln cscu +C secudu = ln secu+tanu +C cscudu = ln cscu+cotu +C 13

14 7.3 Trigonometric Substitution for Integrals Involving a 2 x 2, a 2 +x 2, x 2 a 2 Introduction Consider the integral x a 2 x 2 dx. The substitution of u = a 2 x 2 will allow us to integrate this integral. Now consider a 2 x 2 dx. We have integrated this type of integral using the area of either a half or quarter of a circle depending on the limits of integration. What if we wanted to evaluate this indefinite integral or the definite integral in which the limits do not define either a half or quarter of a circle? We will evaluate this type of integral using a type of substitution called inverse substitution. Trigonometric Substitution for Integrals Involving a 2 x 2, a 2 +x 2, x 2 a 2 Radical Substitution Restrictions on θ Identiy a 2 x 2 x = asinθ π 2 θ π 2 1 sin 2 θ = cos 2 θ a 2 +x 2 x = atanθ π 2 θ π 2 1+tan 2 θ = sec 2 θ x 2 a 2 x = asecθ 0 θ π, θ π 2 sec 2 θ 1 = tan 2 θ Example Evaluate a 2 x 2 dx 14

15 Simplifications of the trigonometric substitutions 1. a 2 x 2 = a 2 a 2 sin 2 θ = a 2 cos 2 θ = acosθ = acosθ 2. a 2 +x 2 = a 2 +a 2 tan 2 θ = a 2 sec 2 θ = asecθ = asecθ 3. x 2 a 2 = a 2 sec 2 θ a 2 = a 2 tan 2 θ = atanθ = atanθ dx Example Evaluate 9+x 2 Example Evaluate 4 2 x 2 4 dx x 15

16 4 x 2 Example Evaluate dx x 2 Example Evaluate dx (5 4x x 2 ) 5/2 16

17 7.4 Integration of Rational Functions by Partial Fractions Reducing an Improper Fraction (long division) When the degree of the numerator is greater than or equal to the degree of the denominator, we will be using long division to simplify the integrand. Sometimes after doing long division, you will have a remainder that will be placed on top of the divisor; this part of the new integrand may result after integration in either a natural logarithm or an arctangent. dx x 2 +a 2 = 1 a tan 1( x ) +c a x 2 +7x 3 Example Evaluate dx x+4 Example Evaluate 5x 5 +28x x 4 +9 dx 17

18 Example Evaluate 3 1 4x 2 7 2x+3 dx Partial Fractions Partial Fractions consists of decomposing a rational function into simpler component fractions and then evaluating the integral term by term. Example Denominator is a product of disctinct linear factors 3x+7 x 2 +6x+5 dx 18

19 Example Denominator is a product of linear factors, some of which are repeated. 3x 2 8x+13 (x+3)(x 1) 2 dx Example Denominator contains irreducible quadratic factors, none of which is repeated. 2x 2 +x 8 x 3 +4x dx 19

20 7.4.3 Rationalizing substitutions Some nonrational functions can be changed into rational functions by means of appropriate substitutions. Example x x 4 dx Additional Examples 2x+1 Example Evaluate x 2 +2x 3 dx 20

21 Example Evaluate x 2 x 21 (x 2 +4)(2x 1) dx Example Evaluate 4x 2 +3x+6 x 2 (x 2 +3) dx 21

22 Example Evaluate e y 16 e 2y dy Example Evaluate dx 1+e x Example Evaluate e 2t e t 2 dt 22

23 x Example Evaluate 2 +x+1dx 23

24 7.5 Strategy for Integration Basic Integration Formulas 1. du = u+c 2. k f(x)dx = k f(x)dx (du±dv) dx = du± dv 4. 1 du = ln u +C 6. u a u du = au +C 8. lna cosudu = sinu+c 10. u n du = un+1 +C; n R, n 1 n+1 e u dx = e u +C sinudu = cosu+c sec 2 udu = tanu+c 11. csc 2 udu = cotu+c 12. secutanudu = secu+c 13. cscucotudu = cscu+c tanudu = cotudu = secudu = cscudu = sinu du = ln cosu +C = ln secu +C cosu cosu du = ln sinu +C = ln cscu +C sinu secu cscu du a 2 u 2 = sin 1 u a +C ( ) secu+tanu du = secu+tanu ( ) cscu+cotu du = cscu+cotu du u u 2 a = 1 u 2 a sec 1 +C a du u 2 a 2 = 1 2a ln u a u+a +C ( sec 2 ) u+secutanu du = ln secu+tanu +C secu+tanu ( csc 2 ) u+cscucotu du = ln cscu+cotu +C cscu+cotu du a 2 +u 2 = 1 a tan 1 u a +C du u 2 ±a 2 = ln u+ u 2 ±a 2 +C 24

25 7.5.2 Procedures for matching integrals to basic formulas 1. Simplify the integrand. 2. Make a simplifying substitution (u-substitution). 3. Compete the square. 4. Use a trigonometric identity / make a trigonometric substitution. 5. Eliminate a square root. 6. Reduce an improper fraction. 7. Separate a fraction / partial fractions. 8. Multiply by a form of Use integration by parts Example xe x2 dx Example xe 2x dx Example x 3 e x2 dx 25

26 Example x 2 6x+9 dx Example x 2 x+4 x 3 +4x dx Example dx x 2 x

27 Example sin 2 xcos 5 xdx Example e x sin(e x ) dx Example dx 6 x 2 +10x dx 27

28 Example π cos(4x)dx Example ( x+10) 3 x dx Example x 5 sec 5 (3x 6 )tan 3 (3x 6 )dx 28

29 7.6 Integral Tables The Basic techniques of integration are substitution and integration by parts; these techniques transform unfamiliar integrals into integrals whose forms are recognizable or can be found in a table. The integral tables were created by applying substitutions and integration by parts to generic integrals in order to save the trouble of repeating laborious calculations. When an integral matches an integral in the table or can be changed into one of the tabulated integrals through algebra, trigonometry substitution or calculus, the tables give a ready made solution for the problem. Example dx x 3x+4 Formula # Example dy 1+9y 2 Formula # 29

30 Example sin(4θ) cot(4θ)dθ Formula # Example x 1 x dx Formula # 30

31 Example dx x 3 3 x 4 Formula # Example x x 2 dx Formula # x 31

32 Example cot 4 (3x)dx Formula # Example dx x 7 x 2 Formula # 32

33 7.7 Approximate Integration NOTE: Approximate/numerical integration is used when either we cannot find an antiderivative to a problem or one does not exist Midpoint Rule A. Formula b f(x)dx M n = a Where x = b a n n f( x i ) x = x[f( x 1 )+f( x 2 )+ +f( x n )] i=1 and x i = x i 1 +x i 2 B. The Error Estimate for the Midpoint Rule, E M. 1. E M = b a = midpoint of [x i 1,x i ]. f(x)dx M n, where M n is the Midpoint Rule. 2. If f is continuous and K is any upper bound for the values of f on [a,b], then ( ) ( ) b a (b a) E M ( x) 2 3 K 24 24n 2 K, where x = b a n Example Use the Midpoint Rule to estimate midpoint approximation, E M. 1 1 (x 2 + 1)dx using n = 4. Find the error in the 33

34 7.7.2 Trapezoidal Rule A. Formula One way to approximate a definite integral is by the use of n trapezoids rather than rectangles. In the development of this method we will assume that the function f(x) is continuous and positive valued on the interval [a,b] and that b and the x-axis, from x = a to x = b. a f(x)dx represents the area of the region bounded by the graph of f(x) Partition the interval [a,b] into n equal subintervals, each of width x = b a such that n a = x 0 < x 1 < x 2 < < x n = b. The area of a trapezoid is A trap = 1 2 h(b 1+b 2 ) where b 1 and b 2 are the two parallel sides of the trapezoid and h is the distance between the two parallel sides. b 2 b 1 h With the trapezoid in this position, the height is h = x and the bases b 1 = f(x 0 ) and b 2 = f(x 1 ). We have formed n trapezoids which have areas given by: ( )( ) 1 b a Area of first trapezoid: A 1 = (f(x 0 )+f(x 1 )) 2 n Area of second trapezoid: A 2 =. Area of n th trapezoid: A n = ( 1 2 ( 1 2 )( b a )( b a n n ) (f(x 1 )+f(x 2 )) ) (f(x n 1 )+f(x n )) When you add up all the areas and combine like terms you get a formula that look like: b a f(x)dx T n = n i=1 A 1 +A 2 + +A n = x 2 [f(x 0)+2f(x 1 )+2f(x 2 )+ +2f(x n 1 )+f(x n )] where x = b a and x i = x 0 +i x. n Note: The coefficients in the Trapezoidal Rule follow the pattern:

35 B. Error Estimate for the Trapezoidal Rule, E T. 1. E T = b a f(x)dx T n, where T n is the Trapezoidal Rule. 2. If f is continuous and K is any upper bound for the values of f on [a,b], then ( ) b a E T ( x) 2 K 12 Example Use the Trapezoidal Rule to estimate ( (b a) 3 12n ) K, where x = b a n (x 2 +1)dx using n = 4. Find the error in the trapezoidal approximation, E T. How large do we have to choose n so that the approximation T n to the integral 1 1 (x 2 +1)dx is accurate to 0.001? 35

36 Example Use the tabulated values of the integrand to estimate the integral the trapezoidal rule. Find the error in the trapezoidal approximation, E T. 3 0 θ dθ using 16+θ 2 θ θ 16+θ Simpson s / Parabolic Rule A. The formula Another way to approximate a definite integral is by the use of n parabolas rather than rectangles or trapezoids. In the development of this method we will assume that the function f(x) is continuous and positive valued on the interval [a,b] and that b the graph of f(x) and the x-axis, from x = a to x = b. a f(x)dx represents the area of the region bounded by Partition the interval [a,b] into n equal subintervals, each of width x = b a such that n a = x 0 < x 1 < x 2 < < x n = b. For Simpson s rule the value of n MUST BE EVEN. 36

37 y 1 y 0 y 2 h h The area under the parabola is A = h 3 (y 0 + 4y 1 + y 2 ). We will use this to estimate the area under a curve. Simpson s Rule (n is even) Let f(x) be continuous on [a, b]. Simpson s Rule for approximating b a b a f(x)dx is given by ( ) b a f(x)dx S n = [f(x 0 )+4f(x 1 )+2f(x 2 )+4f(x 3 )+ +4f(x n 1 )+f(x n )] 3n OR b a f(x)dx S n = ( ) x [f(x 0 )+4f(x 1 )+2f(x 2 )+4f(x 3 )+ +4f(x n 1 )+f(x n )] 3n where x = b a n Note: The coefficients in Simpson s Rule follow the pattern: B. Error Estimate for Simpsons Rule, E S. 1. E s = b a f(x)dx S n, where S n is Simpsons Rule. 2. If f (4) is continuous and K is any upper bound for the values of f (4) on [a,b], then E S ( ) b a ( x) 4 K 180 ( (b a) 3 180n 4 ) K, where x = b a n 37

38 Example Use Simpsons Rule to estimate 1 1 (x 2 +1)dx using n = 4. Find the error in Simpson s approximation, E S. How large do we have to choose n so that the approximation T n to the integral 1 1 (x 2 +1)dx is accurate to ? Example Use the tabulated values of the integrand to estimate the integral Simpson s rule. Find the error in the trapezoidal approximation, E S. 3 0 θ dθ using 16+θ 2 θ θ 16+θ