Trigonometric Integrals: Solving Integrals with Combinations of Trigonometric Functions - , Study notes of Mathematics

Download Trigonometric Integrals: Solving Integrals with Combinations of Trigonometric Functions - and more Study notes Mathematics in PDF only on Docsity! Math 104-003, Fall 2009 Tong Zhu Department of Mathematics University of Pennsylvania September 23, 2009 Tong Zhu Math 104-003, Fall 2009 Trigonometric Integrals Goal: to solve integrals whose integrands are combinations of trigonometric functions. For example,∫ sin4 x cos xdx ∫ sin2 x cos3 xdx∫ sin5 xdx ∫ sin(4x) cos(5x)dx or ∫ tan3 x sec2 xdx ∫ tan3 xdx Methods: We will need I Techniques of Integration: substitution, integral by parts I Trigonometric Identities Tong Zhu Math 104-003, Fall 2009 Trigonometric Integrals Goal: to solve integrals whose integrands are combinations of trigonometric functions. For example,∫ sin4 x cos xdx ∫ sin2 x cos3 xdx∫ sin5 xdx ∫ sin(4x) cos(5x)dx or ∫ tan3 x sec2 xdx ∫ tan3 xdx Methods: We will need I Techniques of Integration: substitution, integral by parts I Trigonometric Identities Tong Zhu Math 104-003, Fall 2009 Trigonometric Integrals Goal: to solve integrals whose integrands are combinations of trigonometric functions. For example,∫ sin4 x cos xdx ∫ sin2 x cos3 xdx∫ sin5 xdx ∫ sin(4x) cos(5x)dx or ∫ tan3 x sec2 xdx ∫ tan3 xdx Methods: We will need I Techniques of Integration: substitution, integral by parts I Trigonometric Identities Tong Zhu Math 104-003, Fall 2009 Trigonometric Integrals Goal: to solve integrals whose integrands are combinations of trigonometric functions. For example,∫ sin4 x cos xdx ∫ sin2 x cos3 xdx∫ sin5 xdx ∫ sin(4x) cos(5x)dx or ∫ tan3 x sec2 xdx ∫ tan3 xdx Methods: We will need I Techniques of Integration: substitution, integral by parts I Trigonometric Identities Tong Zhu Math 104-003, Fall 2009 Some useful trigonometric identities: 1. sin2 x + cos2 x = 1⇒ sin2 x = 1− cos2 x , cos2 x = 1− sin2 x 2. cos(2x) = cos2 x − sin2 x ⇒ sin2 x = 1 2 (1− cos 2x), cos2 x = 1 2 (1 + cos 2x) 3. sin 2x = 2 sin x cos x 4. sec2 x = 1 + tan2 x ⇒ tan2 x = sec2 x − 1 Tong Zhu Math 104-003, Fall 2009 Some useful trigonometric identities: 1. sin2 x + cos2 x = 1⇒ sin2 x = 1− cos2 x , cos2 x = 1− sin2 x 2. cos(2x) = cos2 x − sin2 x ⇒ sin2 x = 1 2 (1− cos 2x), cos2 x = 1 2 (1 + cos 2x) 3. sin 2x = 2 sin x cos x 4. sec2 x = 1 + tan2 x ⇒ tan2 x = sec2 x − 1 Tong Zhu Math 104-003, Fall 2009 Some useful trigonometric identities: 1. sin2 x + cos2 x = 1⇒ sin2 x = 1− cos2 x , cos2 x = 1− sin2 x 2. cos(2x) = cos2 x − sin2 x ⇒ sin2 x = 1 2 (1− cos 2x), cos2 x = 1 2 (1 + cos 2x) 3. sin 2x = 2 sin x cos x 4. sec2 x = 1 + tan2 x ⇒ tan2 x = sec2 x − 1 Tong Zhu Math 104-003, Fall 2009 Example 1 Evaluate ∫ sin4 x cos xdx Solution: Let u = sin x ⇒ du = cos xdx .∫ sin4 x cos xdx = ∫ u4du = 1 5 u5 + C = 1 5 sin5 x + C Tong Zhu Math 104-003, Fall 2009 Example 1 Evaluate ∫ sin4 x cos xdx Solution: Let u = sin x ⇒ du = cos xdx .∫ sin4 x cos xdx = ∫ u4du = 1 5 u5 + C = 1 5 sin5 x + C Tong Zhu Math 104-003, Fall 2009 Example 1 Evaluate ∫ sin4 x cos xdx Solution: Let u = sin x ⇒ du = cos xdx .∫ sin4 x cos xdx = ∫ u4du = 1 5 u5 + C = 1 5 sin5 x + C Tong Zhu Math 104-003, Fall 2009 Example 1 Evaluate ∫ sin4 x cos xdx Solution: Let u = sin x ⇒ du = cos xdx .∫ sin4 x cos xdx = ∫ u4du = 1 5 u5 + C = 1 5 sin5 x + C Tong Zhu Math 104-003, Fall 2009 Example 2 Evaluate ∫ sin3 x cos2 xdx Solution: Still need substitution. But before substitution, write the integral as ∫ sin3 x cos2 xdx = ∫ sin2 x cos2 x sin xdx = ∫ (1−cos2 x) cos2 x sin xdx Let u = cos x ⇒ du = − sin xdx . ∫ (1− cos2 x) cos2 x sin xdx = ∫ (1− u2)u2(−du) = − ∫ [u2 − u4]du = −[1 3 u3 − 1 5 u5] + C = −1 3 cos3 x + 1 5 cos5 x + C Tong Zhu Math 104-003, Fall 2009 Example 2 Evaluate ∫ sin3 x cos2 xdx Solution: Still need substitution. But before substitution, write the integral as ∫ sin3 x cos2 xdx = ∫ sin2 x cos2 x sin xdx = ∫ (1−cos2 x) cos2 x sin xdx Let u = cos x ⇒ du = − sin xdx . ∫ (1− cos2 x) cos2 x sin xdx = ∫ (1− u2)u2(−du) = − ∫ [u2 − u4]du = −[1 3 u3 − 1 5 u5] + C = −1 3 cos3 x + 1 5 cos5 x + C Tong Zhu Math 104-003, Fall 2009 Example 2 Evaluate ∫ sin3 x cos2 xdx Solution: Still need substitution. But before substitution, write the integral as ∫ sin3 x cos2 xdx = ∫ sin2 x cos2 x sin xdx = ∫ (1−cos2 x) cos2 x sin xdx Let u = cos x ⇒ du = − sin xdx . ∫ (1− cos2 x) cos2 x sin xdx = ∫ (1− u2)u2(−du) = − ∫ [u2 − u4]du = −[1 3 u3 − 1 5 u5] + C = −1 3 cos3 x + 1 5 cos5 x + C Tong Zhu Math 104-003, Fall 2009 Example 3 Evaluate ∫ cos4 xdx Solution: Write the integral as∫ cos4 xdx = ∫ [cos2 x ]2dx = ∫ [ 1 2 (1 + cos 2x)]2dx = 1 4 ∫ (1 + 2 cos 2x + cos2 2x)dx = 1 4 [x + 2 · 1 2 sin 2x + ∫ 1 2 (1 + cos 4x)dx ] = 1 4 [x + sin 2x + 1 2 (x + 1 4 sin 4x)] + C = 1 4 [ 3 2 x + sin 2x + 1 8 sin 4x ] + C Tong Zhu Math 104-003, Fall 2009 Example 3 Evaluate ∫ cos4 xdx Solution: Write the integral as∫ cos4 xdx = ∫ [cos2 x ]2dx = ∫ [ 1 2 (1 + cos 2x)]2dx = 1 4 ∫ (1 + 2 cos 2x + cos2 2x)dx = 1 4 [x + 2 · 1 2 sin 2x + ∫ 1 2 (1 + cos 4x)dx ] = 1 4 [x + sin 2x + 1 2 (x + 1 4 sin 4x)] + C = 1 4 [ 3 2 x + sin 2x + 1 8 sin 4x ] + C Tong Zhu Math 104-003, Fall 2009 Example 4 Evaluate ∫ sin2 x cos2 xdx Solution 1 :∫ sin2 x cos2 xdx = ∫ 1 2 (1− cos 2x)1 2 (1 + cos 2x)dx = 1 4 ∫ (1− cos2 2x)dx = 1 4 ∫ [1− 1 2 (1 + cos 4x)]dx = 1 4 [x − 1 2 (x + 1 4 sin 4x)] + C = 1 4 [ 1 2 x − 1 8 sin 4x ] + C = 1 8 x − 1 32 sin 4x + C Tong Zhu Math 104-003, Fall 2009 Solution 2 : ∫ sin2 x cos2 xdx = ∫ 1 4 (2 sin x cos x)2dx = 1 4 ∫ sin2 2xdx = 1 4 ∫ 1 2 (1− cos 4x)dx = 1 8 (1− 1 4 sin 4x) + C = 1 8 x − 1 32 sin 4x + C Tong Zhu Math 104-003, Fall 2009 Solution 2 : ∫ sin2 x cos2 xdx = ∫ 1 4 (2 sin x cos x)2dx = 1 4 ∫ sin2 2xdx = 1 4 ∫ 1 2 (1− cos 4x)dx = 1 8 (1− 1 4 sin 4x) + C = 1 8 x − 1 32 sin 4x + C Tong Zhu Math 104-003, Fall 2009 Case 2 If m is odd, say m = 2k + 1, use sin2 x = 1− cos2 x . Let u = cos x ⇒ du = − sin xdx .∫ sin2k+1 x cosn xdx = ∫ (sin2 x)k cosn x sin xdx = ∫ (1− cos2 x)k cosn x sin xdx = − ∫ (1− u2)kundu Case 3 If both m and n are odd, either Case 1 or Case 2 can be applied. Tong Zhu Math 104-003, Fall 2009 Case 2 If m is odd, say m = 2k + 1, use sin2 x = 1− cos2 x . Let u = cos x ⇒ du = − sin xdx .∫ sin2k+1 x cosn xdx = ∫ (sin2 x)k cosn x sin xdx = ∫ (1− cos2 x)k cosn x sin xdx = − ∫ (1− u2)kundu Case 3 If both m and n are odd, either Case 1 or Case 2 can be applied. Tong Zhu Math 104-003, Fall 2009 Case 4 If both m and n are even, apply sin2 x = 1 2 (1− cos 2x), cos2 x = 1 2 (1 + cos 2x) Sometimes, sin 2x = 2 sin x cos x is useful too. Tong Zhu Math 104-003, Fall 2009 Integrals of Tangents and Secants ∫ tan xdx = ∫ sin x cos x dx = ∫ −1 u du = − ln |u|+C = − ln | cos x |+C (substitution u = cos x) Similarly, ∫ cot xdx = ln | sin x |+ C ∫ sec xdx = ∫ sec x(tan x + sec x) tan x + sec x dx = ∫ sec x tan x + sec2 x tan x + sec x dx Let u = tan x + sec x ⇒ du = (sec2 x + sec x tan x)dx . Tong Zhu Math 104-003, Fall 2009 Integrals of Tangents and Secants ∫ tan xdx = ∫ sin x cos x dx = ∫ −1 u du = − ln |u|+C = − ln | cos x |+C (substitution u = cos x) Similarly, ∫ cot xdx = ln | sin x |+ C ∫ sec xdx = ∫ sec x(tan x + sec x) tan x + sec x dx = ∫ sec x tan x + sec2 x tan x + sec x dx Let u = tan x + sec x ⇒ du = (sec2 x + sec x tan x)dx . Tong Zhu Math 104-003, Fall 2009 Integrals of Tangents and Secants ∫ tan xdx = ∫ sin x cos x dx = ∫ −1 u du = − ln |u|+C = − ln | cos x |+C (substitution u = cos x) Similarly, ∫ cot xdx = ln | sin x |+ C ∫ sec xdx = ∫ sec x(tan x + sec x) tan x + sec x dx = ∫ sec x tan x + sec2 x tan x + sec x dx Let u = tan x + sec x ⇒ du = (sec2 x + sec x tan x)dx . Tong Zhu Math 104-003, Fall 2009 Integrals of Tangents and Secants ∫ tan xdx = ∫ sin x cos x dx = ∫ −1 u du = − ln |u|+C = − ln | cos x |+C (substitution u = cos x) Similarly, ∫ cot xdx = ln | sin x |+ C ∫ sec xdx = ∫ sec x(tan x + sec x) tan x + sec x dx = ∫ sec x tan x + sec2 x tan x + sec x dx Let u = tan x + sec x ⇒ du = (sec2 x + sec x tan x)dx . Tong Zhu Math 104-003, Fall 2009 Integrals of Tangents and Secants ∫ tan xdx = ∫ sin x cos x dx = ∫ −1 u du = − ln |u|+C = − ln | cos x |+C (substitution u = cos x) Similarly, ∫ cot xdx = ln | sin x |+ C ∫ sec xdx = ∫ sec x(tan x + sec x) tan x + sec x dx = ∫ sec x tan x + sec2 x tan x + sec x dx Let u = tan x + sec x ⇒ du = (sec2 x + sec x tan x)dx . Tong Zhu Math 104-003, Fall 2009 Integrals of Tangents and Secants ∫ tan xdx = ∫ sin x cos x dx = ∫ −1 u du = − ln |u|+C = − ln | cos x |+C (substitution u = cos x) Similarly, ∫ cot xdx = ln | sin x |+ C ∫ sec xdx = ∫ sec x(tan x + sec x) tan x + sec x dx = ∫ sec x tan x + sec2 x tan x + sec x dx Let u = tan x + sec x ⇒ du = (sec2 x + sec x tan x)dx . Tong Zhu Math 104-003, Fall 2009 So ∫ sec xdx = ∫ 1 udu = ln |u|+ C = ln | tan x + sec x |+ C . Similarly, ∫ csc xdx = − ln | csc x + cot x |+ C . Generally, an integral of the form ∫ tanm x secn xdx can be computed in the following way: 1. If m is odd, apply tan2 x = sec2 x − 1 and let u = sec x ⇒ du = sec x tan xdx 2. If n is even, apply sec2 x = 1 + tan2 x and let u = tan x ⇒ du = sec2 xdx 3. For other cases, there is no clear-cut guideline. For example,∫ tan xdx and ∫ sec xdx . Tong Zhu Math 104-003, Fall 2009 So ∫ sec xdx = ∫ 1 udu = ln |u|+ C = ln | tan x + sec x |+ C . Similarly, ∫ csc xdx = − ln | csc x + cot x |+ C . Generally, an integral of the form ∫ tanm x secn xdx can be computed in the following way: 1. If m is odd, apply tan2 x = sec2 x − 1 and let u = sec x ⇒ du = sec x tan xdx 2. If n is even, apply sec2 x = 1 + tan2 x and let u = tan x ⇒ du = sec2 xdx 3. For other cases, there is no clear-cut guideline. For example,∫ tan xdx and ∫ sec xdx . Tong Zhu Math 104-003, Fall 2009 So ∫ sec xdx = ∫ 1 udu = ln |u|+ C = ln | tan x + sec x |+ C . Similarly, ∫ csc xdx = − ln | csc x + cot x |+ C . Generally, an integral of the form ∫ tanm x secn xdx can be computed in the following way: 1. If m is odd, apply tan2 x = sec2 x − 1 and let u = sec x ⇒ du = sec x tan xdx 2. If n is even, apply sec2 x = 1 + tan2 x and let u = tan x ⇒ du = sec2 xdx 3. For other cases, there is no clear-cut guideline. For example,∫ tan xdx and ∫ sec xdx . Tong Zhu Math 104-003, Fall 2009 Example 5 Evaluate ∫ tan3 x sec2 xdx Solution 1 : Let u = sec x ⇒ du = sec x tan xdx .∫ tan3 x sec2 xdx = ∫ tan2 x sec x tan x sec xdx = ∫ (u2 − 1)udu = ∫ (u3 − u)du = 1 4 u4 − 1 2 u2 + C = 1 4 sec4 x − 1 2 sec2 x + C Tong Zhu Math 104-003, Fall 2009 Example 5 Evaluate ∫ tan3 x sec2 xdx Solution 1 : Let u = sec x ⇒ du = sec x tan xdx . ∫ tan3 x sec2 xdx = ∫ tan2 x sec x tan x sec xdx = ∫ (u2 − 1)udu = ∫ (u3 − u)du = 1 4 u4 − 1 2 u2 + C = 1 4 sec4 x − 1 2 sec2 x + C Tong Zhu Math 104-003, Fall 2009 Example 5 Evaluate ∫ tan3 x sec2 xdx Solution 1 : Let u = sec x ⇒ du = sec x tan xdx .∫ tan3 x sec2 xdx = ∫ tan2 x sec x tan x sec xdx = ∫ (u2 − 1)udu = ∫ (u3 − u)du = 1 4 u4 − 1 2 u2 + C = 1 4 sec4 x − 1 2 sec2 x + C Tong Zhu Math 104-003, Fall 2009 Solution 2 : Let u = tan x ⇒ du = sec2 xdx .∫ tan3 x sec2 xdx = ∫ u3du = 1 4 u4 + C = 1 4 tan4 x + C Why are these two answers different? 1 4 tan4 x = 1 4 (sec2 x − 1)2 = 1 4 (sec4 x − 2 sec2 x + 1) = 1 4 sec4 x − 1 2 sec2 x + 1 4 So the two answers are the same! Tong Zhu Math 104-003, Fall 2009 Solution 2 : Let u = tan x ⇒ du = sec2 xdx .∫ tan3 x sec2 xdx = ∫ u3du = 1 4 u4 + C = 1 4 tan4 x + C Why are these two answers different? 1 4 tan4 x = 1 4 (sec2 x − 1)2 = 1 4 (sec4 x − 2 sec2 x + 1) = 1 4 sec4 x − 1 2 sec2 x + 1 4 So the two answers are the same! Tong Zhu Math 104-003, Fall 2009