Homework Help is Here – Start Your Trial Now!
Math
Calculus
Step 3 So far we have the following. * sec ¯ 1 (8) cos(0) sin2(0). de We can now use the substitution u = sin(0), so du = E 0 = น 3' √3 2 (cos(0) 63 Further, when 0 = sec sec¯¹(8), u = 8 cos(0) de. Once more, we must also make a substitution for the limits of integration. When 3√√7 8 Step 4 We have determined that if we let u = sin(0), then du = cos(0) de on the interval √3 63 Applying the substitution gives us the following result. 2 8 L sec-1 (8) Jπ/3 r sec ¯1 (8) cos(0) sin2(0) 63/8 1 de = и du 2 3/2 u² Step 5 We can now evaluate the integral. 63/8 3/2 du = 11 du 42 63/8 68 3/2 2 √3 8 3√7 Evaluate the integral. b 8 dx (x² - 1)3/2 Step 1 Recall the Inverse Substitution Rule, where f and g are differentiable functions and g is one-to-one. f(x) dx = = [fg f(g(t))g'(t) dt [F(x) We are given the following. 8 dx (x² - 1)3/2 We note that (x² - 1)3/2. = 3 - 1 Therefore, the following entry from the Table of Trigonometric Substitutions is appropriate. Expression Substitution Identity √√√x²-a² x = a sec(0), 0 ≤ 0 < Π 2 oг π ≤ < 3π 2 sec² (0) - 1 = = tan²(0) 3 3 If - 1 = √x² - a then a = 1 1 Therefore, we can let x = sec(0), so dx = sec tan tan (0) sec (0) ᏧᎾ . Further, when 3 πT We also must make a substitution for the limits of integration in the definite integral. Since x = sec(0), we note that when x = 2, 0 = 3 x = 8,0 = sec 8 8 Step 2 We have determined that if we let x = sec(0), then dx Π = sec(0) tan(0) do on the interval sec ec¯¹(8)]. Applying the substitution gives us the following result. 8 L dx - 1)/3/2 = √ / 3 r sec¯1 (8) sec(0) tan(0) tan³ (0) ᏧᎾ (x²- To evaluate this trigonometric integral, we put everything in terms of sin(0) and cos(0). Doing so gives the following result. * sec¯¹ (8) sec(0) tan(0) tan³ (0) Jπ/3 cos(0) sec¯¹ (8) de = de Jπ/3 sin² (0) sin² (0)

Step 3 So far we have the following. * sec ¯ 1 (8) cos(0) sin2(0). de We can now use the substitution u = sin(0), so du = E 0 = น 3' √3 2 (cos(0) 63 Further, when 0 = sec sec¯¹(8), u = 8 cos(0) de. Once more, we must also make a substitution for the limits of integration. When 3√√7 8 Step 4 We have determined that if we let u = sin(0), then du = cos(0) de on the interval √3 63 Applying the substitution gives us the following result. 2 8 L sec-1 (8) Jπ/3 r sec ¯1 (8) cos(0) sin2(0) 63/8 1 de = и du 2 3/2 u² Step 5 We can now evaluate the integral. 63/8 3/2 du = 11 du 42 63/8 68 3/2 2 √3 8 3√7 Evaluate the integral. b 8 dx (x² - 1)3/2 Step 1 Recall the Inverse Substitution Rule, where f and g are differentiable functions and g is one-to-one. f(x) dx = = [fg f(g(t))g'(t) dt [F(x) We are given the following. 8 dx (x² - 1)3/2 We note that (x² - 1)3/2. = 3 - 1 Therefore, the following entry from the Table of Trigonometric Substitutions is appropriate. Expression Substitution Identity √√√x²-a² x = a sec(0), 0 ≤ 0 < Π 2 oг π ≤ < 3π 2 sec² (0) - 1 = = tan²(0) 3 3 If - 1 = √x² - a then a = 1 1 Therefore, we can let x = sec(0), so dx = sec tan tan (0) sec (0) ᏧᎾ . Further, when 3 πT We also must make a substitution for the limits of integration in the definite integral. Since x = sec(0), we note that when x = 2, 0 = 3 x = 8,0 = sec 8 8 Step 2 We have determined that if we let x = sec(0), then dx Π = sec(0) tan(0) do on the interval sec ec¯¹(8)]. Applying the substitution gives us the following result. 8 L dx - 1)/3/2 = √ / 3 r sec¯1 (8) sec(0) tan(0) tan³ (0) ᏧᎾ (x²- To evaluate this trigonometric integral, we put everything in terms of sin(0) and cos(0). Doing so gives the following result. * sec¯¹ (8) sec(0) tan(0) tan³ (0) Jπ/3 cos(0) sec¯¹ (8) de = de Jπ/3 sin² (0) sin² (0)

Algebra and Trigonometry (MindTap Course List)
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter2: Functions
Section2.4: Average Rate Of Change Of A Function
Problem 4.2E: bThe average rate of change of the linear function f(x)=3x+5 between any two points is ________.
See similar textbooks
icon
Related questions
Question

Please help me with the last part of this problem. I am having trouble understanding what to do. Thank you

Step 3 So far we have the following. * sec ¯ 1 (8) cos(0) sin2(0). de We can now use the substitution u = sin(0), so du = E 0 = น 3' √3 2 (cos(0) 63 Further, when 0 = sec sec¯¹(8), u = 8 cos(0) de. Once more, we must also make a substitution for the limits of integration. When 3√√7 8 Step 4 We have determined that if we let u = sin(0), then du = cos(0) de on the interval √3 63 Applying the substitution gives us the following result. 2 8 L sec-1 (8) Jπ/3 r sec ¯1 (8) cos(0) sin2(0) 63/8 1 de = и du 2 3/2 u² Step 5 We can now evaluate the integral. 63/8 3/2 du = 11 du 42 63/8 68 3/2 2 √3 8 3√7
Transcribed Image Text:Step 3 So far we have the following. * sec ¯ 1 (8) cos(0) sin2(0). de We can now use the substitution u = sin(0), so du = E 0 = น 3' √3 2 (cos(0) 63 Further, when 0 = sec sec¯¹(8), u = 8 cos(0) de. Once more, we must also make a substitution for the limits of integration. When 3√√7 8 Step 4 We have determined that if we let u = sin(0), then du = cos(0) de on the interval √3 63 Applying the substitution gives us the following result. 2 8 L sec-1 (8) Jπ/3 r sec ¯1 (8) cos(0) sin2(0) 63/8 1 de = и du 2 3/2 u² Step 5 We can now evaluate the integral. 63/8 3/2 du = 11 du 42 63/8 68 3/2 2 √3 8 3√7
Evaluate the integral. b 8 dx (x² - 1)3/2 Step 1 Recall the Inverse Substitution Rule, where f and g are differentiable functions and g is one-to-one. f(x) dx = = [fg f(g(t))g'(t) dt [F(x) We are given the following. 8 dx (x² - 1)3/2 We note that (x² - 1)3/2. = 3 - 1 Therefore, the following entry from the Table of Trigonometric Substitutions is appropriate. Expression Substitution Identity √√√x²-a² x = a sec(0), 0 ≤ 0 < Π 2 oг π ≤ < 3π 2 sec² (0) - 1 = = tan²(0) 3 3 If - 1 = √x² - a then a = 1 1 Therefore, we can let x = sec(0), so dx = sec tan tan (0) sec (0) ᏧᎾ . Further, when 3 πT We also must make a substitution for the limits of integration in the definite integral. Since x = sec(0), we note that when x = 2, 0 = 3 x = 8,0 = sec 8 8 Step 2 We have determined that if we let x = sec(0), then dx Π = sec(0) tan(0) do on the interval sec ec¯¹(8)]. Applying the substitution gives us the following result. 8 L dx - 1)/3/2 = √ / 3 r sec¯1 (8) sec(0) tan(0) tan³ (0) ᏧᎾ (x²- To evaluate this trigonometric integral, we put everything in terms of sin(0) and cos(0). Doing so gives the following result. * sec¯¹ (8) sec(0) tan(0) tan³ (0) Jπ/3 cos(0) sec¯¹ (8) de = de Jπ/3 sin² (0) sin² (0)
Transcribed Image Text:Evaluate the integral. b 8 dx (x² - 1)3/2 Step 1 Recall the Inverse Substitution Rule, where f and g are differentiable functions and g is one-to-one. f(x) dx = = [fg f(g(t))g'(t) dt [F(x) We are given the following. 8 dx (x² - 1)3/2 We note that (x² - 1)3/2. = 3 - 1 Therefore, the following entry from the Table of Trigonometric Substitutions is appropriate. Expression Substitution Identity √√√x²-a² x = a sec(0), 0 ≤ 0 < Π 2 oг π ≤ < 3π 2 sec² (0) - 1 = = tan²(0) 3 3 If - 1 = √x² - a then a = 1 1 Therefore, we can let x = sec(0), so dx = sec tan tan (0) sec (0) ᏧᎾ . Further, when 3 πT We also must make a substitution for the limits of integration in the definite integral. Since x = sec(0), we note that when x = 2, 0 = 3 x = 8,0 = sec 8 8 Step 2 We have determined that if we let x = sec(0), then dx Π = sec(0) tan(0) do on the interval sec ec¯¹(8)]. Applying the substitution gives us the following result. 8 L dx - 1)/3/2 = √ / 3 r sec¯1 (8) sec(0) tan(0) tan³ (0) ᏧᎾ (x²- To evaluate this trigonometric integral, we put everything in terms of sin(0) and cos(0). Doing so gives the following result. * sec¯¹ (8) sec(0) tan(0) tan³ (0) Jπ/3 cos(0) sec¯¹ (8) de = de Jπ/3 sin² (0) sin² (0)
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer