Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Section 6.2 Trigonometric Integrals and Substitutions 2010 Kiryl Tsishchanka Trigonometric Integrals and Substitutions This section consists of two parts: Part I: Trigonometric Integrals. We will distinguish three main cases: Case A: Integrals of type Z sinm x cosn xdx where m and n are nonnegative integers. METHOD OF INTEGRATION: (i) If m is odd, then u = cos x. (ii) If n is odd, then u = sin x. (iii) If both m and n are even, then use the identities 1 sin2 x = (1 − cos 2x), 2 1 cos2 x = (1 + cos 2x) 2 or(and) sometimes sin x cos x = EXAMPLE 1: Find Z 1 sin 2x 2 sin3 xdx. Solution: We have Z 3 sin xdx = =− Z EXAMPLE 2: Find Z 2 sin x · sin xdx = Z d cos x = du (1 − cos x) sin xdx = − sin xdx = du 2 1 1 (1 − u2 )du = −u + u3 + C = − cos x + cos3 x + C 3 3 Z cos x = u sin5 xdx. 1 sin xdx = −du Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 2: Find Z 2010 Kiryl Tsishchanka sin5 xdx. Solution: We have Z Z Z Z 5 4 2 2 sin xdx = sin x · sin xdx = (sin x) sin xdx = (1 − cos2 x)2 sin xdx cos x = u d cos x = du = − sin xdx = du Z Z = − (1 − u2 )2 du = − (1 − 2u2 + u4 )du sin xdx = −du 2 1 2 1 = −u + u3 − u5 + C = − cos x + cos3 x − cos5 x + C 3 5 3 5 EXAMPLE 3: Find Z sin2 x cos3 xdx. Solution: We have Z Z Z 2 3 2 2 sin x cos xdx = sin x cos x · cos xdx = sin2 x(1 − sin2 x) cos xdx sin x = u = d sin x = du = cos xdx = du EXAMPLE 4: Find Z √ 3 Z 2 2 u (1 − u )du = Z 1 1 1 1 (u2 − u4 )du = u3 − u5 + C = sin3 x − sin5 x + C 3 5 3 5 sin x cos5 xdx. 2 Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 4: Find Z √ 3 2010 Kiryl Tsishchanka sin x cos5 xdx. Solution: We have Z √ Z √ Z √ 3 3 3 5 4 sin x cos xdx = sin x cos x · cos xdx = sin x(cos2 x)2 · cos xdx sin x = u Z √ Z √ 3 2 2 3 = sin x(1 − sin x) · cos xdx = d sin x = du = u(1 − u2 )2 du cos xdx = du = Z u = Z (u1/3 − 2u7/3 + u13/3 )du = 1/3 2 4 (1 − 2u + u )du = Z = (u1/3 − 2u2+1/3 + u4+1/3 )du u7/3+1 u13/3+1 u1/3+1 −2 + +C 1/3 + 1 7/3 + 1 13/3 + 1 u4/3 u10/3 u16/3 −2 + +C 4/3 10/3 16/3 3 3 3 = u4/3 − u10/3 + u16/3 + C 4 5 16 = EXAMPLE 5: Find Z 3 3 3 4/3 sin x − sin10/3 x + sin16/3 x + C 4 5 16 sin2 x cos2 xdx. Solution: We have 2 Z Z Z Z Z 1 1 1 1 2 2 2 2 sin x cos xdx = (sin x cos x) dx = sin 2xdx = sin 2x dx = (1 − cos 4x)dx 2 4 4 2 4x = u d(4x) = du Z Z 1 1 1 1 (1 − cos 4x)dx = 4dx = du = (1 − cos u)du = (u − sin u) + C = (4x − sin 4x) + C = 8 32 32 32 1 dx = du 4 EXAMPLE 6: Find Z cos4 xdx. 3 Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 6: Find Z 2010 Kiryl Tsishchanka cos4 xdx. Solution: We have 2 Z Z Z Z 1 + cos 2x 1 4 2 2 cos xdx = (cos x) dx = (1 + cos 2x)2 dx dx = 2 4 Z Z 1 1 1 2 = (1 + 2 cos 2x + cos 2x)dx = 1 + 2 cos 2x + (1 + cos 4x) dx 4 4 2 Z 1 3 1 1 3 1 1 1 1 1 = + 2 cos 2x + cos 4x dx = · x + · 2 · sin 2x + · · sin 4x + C 4 2 2 4 2 4 2 4 2 4 1 1 3 sin 4x + C = x + sin 2x + 8 4 32 Case B: Integrals of type Z tanm x secn xdx Midterms where m and n are nonnegative integers. METHOD OF INTEGRATION: (i) If m is odd, then u = sec x. (ii) If n is even, then u = tan x. (iii) In other cases the guidelines are not as clear-cut. EXAMPLE 7: Find Z tan xdx. Solution 1: We have Z tan xdx = Z sin x dx = cos x Z cos x = u d cos x = du 1 · sin xdx = − sin xdx = du cos x sin xdx = −du Z = − 1 du u = − ln |u| + C = − ln | cos x| + C = ln | cos x|−1 + C = ln | sec x| + C Solution 2: We have Z tan xdx = Z tan x sec x dx = sec x sec x = u d sec x = du sec x tan xdx = du 4 = Z 1 du = ln |u| + C = ln | sec x| + C u Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 8: Find Z (a) csc xdx 2010 Kiryl Tsishchanka Z (b) sec xdx Solution 1: (a) We have x =u 2 x Z d Z Z Z Z Z = du dx dx 2du du 2du 2 x = csc xdx = = = = = sin x sin (2u) 2 sin u cos u sin u cos u 1 sin 2 · dx = du 2 2 dx = 2du tan u = v Z Z Z Z Z sec2 udu sec2 udu sec2 udu dv sec2 udu d tan u = dv = = = = = = sin u cos u sin u sec2 u sin u cos u tan u v cos2 u cos u sec2 udu = dv x = ln |v| + C = ln | tan u| + C = ln tan +C 2 In short, 2 x Z Z Z Z Z sec dx dx dx dx 2 x csc xdx = = x = x x = x x sin x 2 cos sin cos sin 2 · 2 sin 2 sec 2 2 2 2 2 2 x =v tan 2 x x Z sec2 x = dv Z dv dx d tan 2 2 = = ln |v| + C = ln tan = +C x = 1 x sec2 v 2 2 tan dx = dv 2 2 2 x 2 dx = 2dv sec 2 This can be rewritten as ln |csc x − cot x| + C (see Appendix II). (b) We have Z sec xdx = Z dx = cos x Z = π sin x + 2 dx π x+ =u 2 π = du d x+ 2 dx = du Z du = sin u u π which is ln tan + C by (a). Substituting x + for u, we get 2 2 π Z x+ 2 + C = ln tan x + π + C sec xdx = ln tan 2 2 4 This can be rewritten as ln |sec x + tan x| + C (see Appendix II). 5 Section 6.2 Trigonometric Integrals and Substitutions 2010 Kiryl Tsishchanka Solution 2: (a) We have Z csc xdx = 1 dx = sin x Z =− Z sin x dx = sin2 x Z 1 du = − (1 − u)(1 + u) Z Z cos x = u d cos x = du sin x dx = − sin xdx = du 1 − cos2 x 1 2 Z 1 =− du 1 − u2 sin xdx = −du Z Z 1 1 1 1 1 1 du = − + du − du 1−u 1+u 2 1−u 2 1+u 1 1 1 1 1 − u 1 − cos x = ln |1 − u| − ln |1 + u| + C = ln + C = ln +C 2 2 2 1 + u 2 1 + cos x This can be rewritten as ln |csc x − cot x| + C (see Appendix II). (b) We have Z cos x dx = cos2 x Z 1 2 sec xdx = Z 1 dx = cos x = Z 1 du = (1 − u)(1 + u) Z Z sin x = u cos x 2 dx = d sin x = du = 1 − sin x cos xdx = du 1 1 + 1−u 1+u 1 du = 2 Z Z 1 1 du + 1−u 2 1 du 1 − u2 Z 1 du 1+u 1 1 1 1 + u 1 + sin x 1 + C = ln +C = − ln |1 − u| + ln |1 + u| + C = ln 2 2 2 1 − u 2 1 − sin x This can be rewritten as ln |sec x + tan x| + C (see Appendix II). Solution 3: (a) We have Z csc xdx = = Z Z csc x csc x − cot x dx = csc x − cot x Z csc x − cot x = u csc2 x − csc x cot x dx = csc x − cot x d(csc x − cot x) = du (− csc x cot x + csc2 x)dx = du 1 du = ln |u| + C = ln | csc x − cot x| + C u (b) We have Z sec xdx = Z sec x sec x + tan x dx = sec x + tan x Z 2 sec x + sec x tan x dx = sec x + tan x 1 du = ln |u| + C = ln | sec x + tan x| + C u Z EXAMPLE 9: Find tan8 x sec4 xdx. = Z 6 sec x + tan x = u d(sec x + tan x) = du (sec x tan x + sec2 x)dx = du Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 9: Find Solution: We have Z Z tan8 x sec4 xdx = = Z tan8 x sec4 xdx. Z u8 (1 + u2)du = EXAMPLE 10: Find Z 2010 Kiryl Tsishchanka Z tan8 x sec2 x · sec2 xdx = Z tan x = u tan8 x(1 + tan2 x) sec2 xdx = d tan x = du sec2 xdx = du 1 1 1 1 (u8 + u10 )du = u9 + u11 + C = tan9 x + tan11 x + C 9 11 9 11 tan3 xdx. Solution 1: We have Z Z Z 3 2 tan xdx = tan x · tan xdx = (sec2 x − 1) · tan xdx sec x = u Z Z (u2 − 1)du (sec2 x − 1) · sec x tan xdx d sec x = du = = = sec x u sec x tan xdx = du Z 1 u2 1 = u− du = − ln |u| + C = sec2 x − ln | sec x| + C u 2 2 Solution 2: We have Z Z Z Z 3 2 2 tan xdx = tan x · tan xdx = (sec x − 1) · tan xdx = (sec2 x tan x − tan x)dx = To evaluate Z Z Z sec x tan xdx − Z tan xdx = [by Example 7] = Z sec2 x tan xdx − ln | sec x| Z udu = sec2 x tan xdx, we either 2 sec x tan xdx = Z or Z Therefore 2 sec x · sec x tan xdx = 2 sec x = u d sec x = du sec x tan xdx = du tan x = u sec x tan xdx = d tan x = du = sec2 xdx = du Z udu = = 1 u2 + C = sec2 x + C 2 2 1 u2 + C = tan2 x + C 2 2 1 sec2 x − ln | sec x| + C 2 or 1 tan2 x − ln | sec x| + C 2 Other possible answers are Z 1 tan3 xdx = sec2 x + ln | cos x| + C 2 or 1 tan2 x + ln | cos x| + C 2 Z tan3 xdx = 7 Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 11: Find Z 2010 Kiryl Tsishchanka sec3 xdx. Solution: We have Z 3 sec xdx = Z 2 sec x · sec xdx = sec x = u sec2 xdx = dv d(sec x) = du tan x = v It follows that 2 so 1 sec xdx = 2 3 = sec x tan x − Z sec x tan2 xdx sec x tan xdx = du Z Z 2 = sec x tan x − sec x(sec x − 1)dx = sec x tan x − (sec3 x − sec x)dx = sec x tan x − Z Z Z 3 sec xdx + 3 Z sec xdx = sec x tan x + sec xdx Z sec xdx Z 1 sec x tan x + sec xdx = [by Example 8] = (sec x tan x + ln | sec x + tan x|) + C 2 Case C: Integrals of type Z sin αx sin βxdx, Z sin αx cos βxdx or Z cos αx cos βxdx METHOD OF INTEGRATION: Use the identities 1 sin A sin B = [cos(A − B) − cos(A + B)] 2 1 cos A cos B = [cos(A + B) + cos(A − B)] 2 1 sin A cos B = [sin(A + B) + sin(A − B)] 2 EXAMPLE 12: Find Z cos 3x cos 2xdx. 8 Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 12: Find Z Solution 1: We have Z 2010 Kiryl Tsishchanka cos 3x cos 2xdx. cos 3x cos 2xdx = 1 cos A cos B = [cos(A + B) + cos(A − B)] 2 1 [cos(3x + 2x) + cos(3x − 2x)]dx 2 Z Z 1 1 cos 5xdx + cos xdx = 2 2 = Z = 1 1 sin 5x + sin x + C 10 2 Solution 2: We have Z cos 3x = u cos 2xdx = dv 1 sin 2x = v 2 cos 3x cos 2xdx = d(cos 3x) = du −3 sin 3xdx = du 1 = cos 3x · sin 2x − 2 3 1 = cos 3x sin 2x + 2 2 1 sin 2x · (−3) sin 3xdx 2 Z Z sin 2x sin 3xdx sin 2xdx = dv 1 − cos 2x = v 2 sin 3x = u = d(sin 3x) = du 3 cos 3xdx = du Z 1 1 1 3 = cos 3x sin 2x + sin 3x · − cos 2x − − cos 2x · 3 cos 3xdx 2 2 2 2 1 3 9 = cos 3x sin 2x − sin 3x cos 2x + 2 4 4 therefore 5 − 4 hence Z Z Z cos 2x cos 3xdx cos 3x cos 2xdx = 1 3 cos 3x sin 2x − sin 3x cos 2x + C1 2 4 1 3 cos 3x sin 2x − sin 3x cos 2x + C1 2 4 4 cos 3x cos 2xdx = − 5 3 2 = − cos 3x sin 2x + sin 3x cos 2x + C 5 5 9 Section 6.2 Trigonometric Integrals and Substitutions 2010 Kiryl Tsishchanka Part II: Trigonometric Substitutions. Here we deal with integrals that involve √ √ √ a2 − x2 , a2 + x2 , x2 − a2 METHOD OF INTEGRATION: √ π π (i) If a2 − x2 , then x = a sin θ, − ≤ θ ≤ . 2 2 √ π π (ii) If a2 + x2 , then x = a tan θ, − < θ < . 2 2 √ 3π π (iii) If x2 − a2 , then x = a sec θ, 0 ≤ θ < or π ≤ θ < . 2 2 Z √ EXAMPLE 13: Find 1 − x2 dx. Solution 1: We have Z √ 1 − x2 dx = Z √ Solution 2: We have √ 1− dx = d(sin θ) p √ sin 2θ = 2 sin θ cos θ = 2x 1 − x2 Therefore x2 dx x = sin θ dx = cos θdθ √ √ 1 − x2 = 1 − sin2 θ = cos2 θ = | cos θ| = cos θ Z Z Z 1 1 1 2 (1 + cos 2θ)dθ = θ + sin 2θ + C = cos θ · cos θdθ = cos θdθ = 2 2 4 Note that Z √ 1 − x2 = u 1 − x2 dx = √ 2 = d( 1 − x ) = du −x √ dx = du 1 − x2 dx = dv 1 √ 1 −1 sin x + x 1 − x2 + C 2 2 Z x = v = √1 − x2 · x − x · √ −x dx 1 − x2 Z Z √ √ 1 − x2 − 1 1 − x2 1 −x2 2 2 √ √ √ dx = x 1 − x − dx = x 1 − x − −√ dx 1 − x2 1 − x2 1 − x2 1 − x2 Z √ Z √ Z √ √ 1 1 − x2 dx + √ 1 − x2 dx + sin−1 x dx = x 1 − x2 − = x 1 − x2 − 2 1−x Therefore Z √ √ 1 − x2 dx = x 1 − x2 + sin−1 x + C1 2 √ = x 1 − x2 − so Z Z √ 1 √ 1 1 − x2 dx = x 1 − x2 + sin−1 x + C 2 2 10 Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 14: Find Z √ 2010 Kiryl Tsishchanka 1 + x2 dx. Solution: We have Z √ x = tan θ dx = d(tan θ) 1 + x2 dx = dx = sec2 θdθ √ √ √ 1 + x2 = 1 + tan2 θ = sec2 θ = | sec θ| = sec θ Z Z 1 2 = sec θ · sec θdθ = sec3 θdθ = [Example 11] = (sec θ tan θ + ln | sec θ + tan θ|) + C 2 √ 1 √ = (x 1 + x2 + ln(x + 1 + x2 )) + C 2 EXAMPLE 15: Find Z √ x2 − 1dx. Solution: We have Z √ x2 x = sec θ − 1dx = dx = d(sec θ) dx = sec θ tan θdθ √ x2 − 1 = sec2 θ − 1 = tan2 θ = | tan θ| = tan θ Z Z = tan θ · sec θ tan θdθ = tan2 θ sec θdθ = = Z Z √ √ 2 (sec θ − 1) sec θdθ = 3 sec θdθ − Z Z (sec3 θ − sec θ)dθ sec θdθ = [Examples 8 and 11] 1 = (sec θ tan θ + ln | sec θ + tan θ|) − ln | sec θ + tan θ| + C 2 = 1 1 sec θ tan θ + ln | sec θ + tan θ| − ln | sec θ + tan θ| + C 2 2 = 1 1 sec θ tan θ − ln | sec θ + tan θ| + C 2 2 1 = (sec θ tan θ − ln | sec θ + tan θ|) + C 2 √ 1 √ = (x x2 − 1 − ln |x + x2 − 1|) + C 2 11 Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 16: Find Z √ 2010 Kiryl Tsishchanka 4 − x2 dx. x2 Solution 1: We have Z √ 4− x2 x2 dx = x = 2 sin θ dx = d(2 sin θ) dx = 2 cos θdθ p p √ 4 − x2 = 4 − 4 sin2 θ = 4(1 − sin2 θ) = 4 cos2 θ = 2| cos θ| = 2 cos θ Z Z Z Z 2 cos θ sin2 θ cos2 θ 1 − sin2 θ 1 = · 2 cos θdθ = dθ = dθ = − dθ 4 sin2 θ sin2 θ sin2 θ sin2 θ sin2 θ √ = Note that cot θ = Therefore Z √ Solution 2: We have √ √ √ Z (csc2 θ − 1)dθ = − cot θ − θ + C 4 − x2 x 4 − x2 4 − x2 −1 x +C dx = − − sin x2 x 2 !! 2!!! ! !! ! θ ! √ x 4 − x2 1 dx = dv 4− =u x2 Z Z √ √ 1 4 − x2 1 −x 1 √ 2 2 √ − − dx = d( 4 − x ) = du − = v = 4 − x · − dx 2 x x2 x x 4 − x −x √ dx = du 2 4−x √ √ Z x 4 − x2 4 − x2 dx +C − √ =− − sin−1 =− x x 2 4 − x2 EXAMPLE 17: Find Z x2 x2 √ 1 dx. x2 − 9 12 Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 17: Find Z x2 √ 2010 Kiryl Tsishchanka 1 dx. x2 − 9 Solution 1: We have Z 1 √ dx = x2 x2 − 9 Note that Therefore x = 3 sec θ dx = d(3 sec θ) dx = 3 sec θ tan θdθ √ p x2 − 9 = 9 sec2 θ − 9 = 9(sec2 θ − 1) = 9 tan2 θ = 3| tan θ| = 3 tan θ Z Z 1 1 1 = cos θdθ = sin θ + C · 3 sec θ tan θdθ = 2 9 sec θ · 3 tan θ 9 9 ! x !!! √ √ ! x2 − 9 x2 − 9 !! ! sin θ = !! θ x 3 √ Z 2 x −9 1 1 √ +C dx = · 2 2 9 x x x −9 √ √ Solution 2: We first note that √ Z √ 2 Z √ 2 Z x −9 x −9 x2 − 9 1 1 √ dx = − dx dx = x2 (x2 − 9) 9 x2 − 9 x2 x2 x2 − 9 √ Z x2 − 9 1 1 √ dx − = 9 x2 x2 − 9 Z Z √ 2 1 x −9 1 1 √ = dx − dx 2 9 9 x2 x −9 We have √ 1 dx = dv x2 − 9 = u x2 Z √ 2 Z √ √ x −9 1 1 2 −9· d( x2 − 9) = du − = v dx = dx = x x2 x2 x x √ dx = du x2 − 9 Z √ 1 x 1 √ − − dx = x2 − 9 − 2 x x x −9 √ Z x2 − 9 1 =− + √ dx 2 x x −9 Therefore Z Z √ 2 Z x −9 1 1 1 1 √ √ dx dx = dx − 2 9 9 x2 x x2 − 9 x2 − 9 √ √ Z Z x2 − 9 1 x2 − 9 1 1 1 1 1 √ √ dx + · dx = · − +C = 9 9 x 9 9 x x2 − 9 x2 − 9 13 Section 6.2 Trigonometric Integrals and Substitutions Solution 3 (version 1): Note that Z 2010 Kiryl Tsishchanka 1 √ dx can be rewritten as 2 x x2 − 9 Z xm (a + bxn )p dx with 1 m = −2, n = 2, p = − , a = −9, and b = 1 2 It is known that if m+1 + p is an integer (which is exactly the case) n then b+ a = uN , where N is the denominator of p xn Therefore we have Z 1 √ dx = 2 2 x x −9 1 = 9 9 1 − 2 = u2 =⇒ x 9 d 1 − 2 = du2 x r 9 1 − 2 = u =⇒ x √ x2 − 9 =u x 18 dx = 2udu x3 1 1 x 1 1 1 1 1 √ · dx = du =⇒ 3 · √ dx = du =⇒ dx = du 3 x u 9 x 9 9 x2 − 9 x2 x2 − 9 Z 1 1 du = u + C = · 9 9 Solution 3 (version 2): We have √ √ x2 − 9 +C x x2 − 9 =u x √ 2 x −9 = du d x Z 1 √ dx = 2 2 x x −9 √9 dx = du 2 2 x x −9 1 1 √ dx = du 9 x2 x2 − 9 √ 1Z x2 − 9 1 1 du = u + C = · +C = 9 9 9 x 14 Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 18: Find √ 3Z 3/2 2010 Kiryl Tsishchanka x3 dx. (4x2 + 9)3/2 0 Solution: We have √ 3Z 3/2 3 x dx = (4x2 + 9)3/2 0 Since √ 3 x = tan θ 2 3 dx = d tan θ 2 3 dx = sec2 θdθ 2 s r 2 √ √ 3 9 4x2 + 9 = 4 tan θ + 9 = 4 · tan2 θ + 9 = 9 tan2 θ + 9 2 4 p √ = 9(tan2 θ + 1) = 9 sec2 θ = 3| sec θ| = 3 sec θ 4x2 + 9 = 3 sec θ, we get √ (4x2 + 9)3/2 = ( 4x2 + 9)3 = (3 sec θ)3 = 27 sec3 θ We also note that since x = 3 tan θ, we have 2 if x = 0, then θ = 0 √ if x = 3 3/2, then θ = π/3 and Therefore √ 3Z 3/2 x3 dx = (4x2 + 9)3/2 0 3 = 16 tan3 θ 3 3 8 · sec2 θdθ = 3 27 sec θ 2 16 0 Zπ/3 0 3 = 16 Zπ/3 27 Zπ/3 tan3 θ 3 dθ = sec θ 16 0 cos θ = u 1 − cos2 θ d(cos θ) = du = − 3 sin θdθ = 2 cos θ 16 − sin θdθ = du EXAMPLE 19: Find Z √ x dx. x2 − 9 15 sin3 θ 3 dθ = 2 cos θ 16 0 Z1/2 1 − u2 3 du = − 2 u 16 1 1/2 Z1/2 3 1 1 3 3 −2 = u+ + 2 − (1 + 1) = (1 − u )du = 16 u 1 16 2 32 1 Zπ/3 Zπ/3 0 Z1/2 (u−2 − 1)du 1 sin2 θ · sin θdθ cos2 θ Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE 19: Find Z √ 2010 Kiryl Tsishchanka x dx. −9 x2 Solution 1: We have Z x2 − 9 = u d(x2 − 9) = du x √ dx = 2xdx = du x2 − 9 1 xdx = du 2 Z Z 1 du 1 u−1/2+1 1 −1/2 = √ u du = = · +C 2 u 2 2 −1/2 + 1 = √ u+C = √ x2 − 9 + C Solution 2: We have Z x √ dx = x2 − 9 dx = d(3 sec θ) dx = 3 sec θ tan θdθ √ p x2 − 9 = 9 sec2 θ − 9 = 9(sec2 θ − 1) = 9 tan2 θ = 3| tan θ| = 3 tan θ Z Z 3 sec θ · 3 sec θ tan θdθ = 3 sec2 θdθ = 3 tan θ + C = 3 tan θ ! x !! √ ! √ ! x2 − 9 x2 − 9 !! ! tan θ = !! θ 3 3 √ Z x x2 − 9 √ 2 √ = x −9+C dx = 3 3 x2 − 9 Note that Therefore EXAMPLE 20: Find Solution: We have Z x = 3 sec θ Z √ 1 x2 −9 dx. x = 3 sec θ √ dx = x2 − 9 1 √ √ dx = d(3 sec θ) dx = 3 sec θ tan θdθ √ p x2 − 9 = 9 sec2 θ − 9 = 9(sec2 θ − 1) = 9 tan2 θ = 3| tan θ| = 3 tan θ Z Z 1 · 3 sec θ tan θdθ = sec θdθ = [Example 8] = ln | sec θ + tan θ| + C1 = 3 tan θ √ x x2 − 9 Note that sec θ = and tan θ = , therefore 3 3 √ Z 2 − 9 √ x 1 x + C1 = ln |x + x2 − 9| + C √ dx = ln | sec θ + tan θ| + C1 = ln + 3 3 x2 − 9 √ √ 16 Section 6.2 Trigonometric Integrals and Substitutions 2010 Kiryl Tsishchanka Appendix I EXAMPLE: Find Z sin3 xdx. Solution 1: We have " Z Z sin3 xdx = sin2 x · sin xdx = sin2 x = u sin xdx = dv 2 sin x cos xdx = du − cos x = v 2 = sin x(− cos x) − 2 = − sin x cos x + 2 2 = − sin x cos x + 2 2 = − sin x cos x + 2 2 = − sin x cos x + 2 Z Z Z Z Z (− cos x) · 2 sin x cos xdx sin x cos2 xdx sin x(1 − sin2 x)dx (sin x − sin3 x)dx sin xdx − 2 2 = − sin x cos x − 2 cos x − 2 therefore 3 so Z # Z Z sin3 xdx sin3 xdx sin3 xdx = − sin2 x cos x − 2 cos x + C1 1 2 sin3 xdx = − sin2 x cos x − cos x + C 3 3 Z Solution 2: We have Z 3 sin xdx = =− Z Z 2 sin x · sin xdx = Z cos x = u d cos x = du (1 − cos x) sin xdx = − sin xdx = du 2 1 1 (1 − u2 )du = −u + u3 + C = − cos x + cos3 x + C 3 3 17 sin xdx = −du Section 6.2 Trigonometric Integrals and Substitutions EXAMPLE: Find Z 2010 Kiryl Tsishchanka sinn xdx. Solution: We have " Z Z sinn xdx = sinn−1 x · sin xdx = sinn−1 x = u sin xdx = dv (n − 1) sinn−2 x cos xdx = du − cos x = v = sin therefore n so Z n−1 x(− cos x) − = − sin n−1 = − sin n−1 = − sin n−1 = − sin n−1 n sin xdx = − sin n−1 Z # (− cos x) · (n − 1) sinn−2 x cos xdx x cos x + (n − 1) x cos x + (n − 1) x cos x + (n − 1) Z Z Z sinn−2 x cos2 xdx sinn−2 x(1 − sin2 x)dx (sinn−2 x − sinn x)dx x cos x + (n − 1) Z sin x cos x + (n − 1) Z sinn−2 xdx n−2 xdx − (n − 1) Z sinn xdx Z n−1 1 n−1 sinn−2 xdx x cos x + sin xdx = − sin n n In particular, if n = 4 we have Z Z 1 3 3 4 sin xdx = − sin x cos x + sin2 xdx 4 4 Z 1 3 1 3 1 = − sin x cos x + − sin x cos x + dx 4 4 2 2 Z n 3 3 1 = − sin3 x cos x − sin x cos x + x + C 4 8 8 Alternatively, Z Z 1 − cos 2x 2 2 Z 1 sin xdx = (sin x) dx = (1 − cos 2x)2 dx dx = 4 Z Z 1 1 1 2 = (1 − 2 cos 2x + cos 2x)dx = 1 − 2 cos 2x + (1 + cos 4x) dx 4 4 2 Z 3 1 3 1 1 1 − 2 cos 2x + cos 4x dx = x − sin 2x + sin 4x + C = 4 2 2 8 4 32 4 Z 2 2 18 Section 6.2 Trigonometric Integrals and Substitutions 2010 Kiryl Tsishchanka Similarly, if n = 5 we have Z Z 1 4 4 5 sin xdx = − sin x cos x + sin3 xdx 5 5 1 4 1 4 = − sin x cos x + − sin2 x cos x + 5 5 3 1 4 1 − sin2 x cos x − = − sin4 x cos x + 5 5 3 2 3 Z sin xdx 2 cos +C1 3 1 4 8 = − sin4 x cos x − sin2 x cos x − cos x + C 5 15 15 Alternatively, Z Z Z 5 4 sin xdx = sin x · sin xdx = (sin2 x)2 · sin xdx = Z cos x = u d cos x = du (1 − cos x) sin xdx = − sin xdx = du 2 2 sin xdx = −du =− Z Z = − (1 − u2 )2 du 2 1 2 1 (1 − 2u2 + u4 )du = −u + u3 − u5 + C = − cos x + cos3 x − cos5 x + C 3 5 3 5 19 Section 6.2 Trigonometric Integrals and Substitutions 2010 Kiryl Tsishchanka Appendix II x 1 1 − cos x cos x = ln 1. ln tan − = ln sin x sin x = ln |csc x − cot x| 2 sin x π π x π 1 − cos x + x+ 1 + sin x 1 sin x 2 = ln 2 = ln = ln 2. ln tan + + = ln tan cos x cos x cos x π 2 4 2 sin x + 2 = ln |sec x + tan x| 1 3. ln 2 1 − cos x 1 + cos x 1 = ln 2 (1 − cos x)2 (1 − cos x)(1 + cos x) 1 (1 − cos x)2 (1 − cos x)2 = ln 1 − cos2 x 2 sin2 x 2 1 − cos x 1 − cos x 1 1 − cos x 1 = ln = ln = · 2 ln sin x 2 sin x 2 sin x 1 cos x = ln |csc x − cot x| − = ln sin x sin x 1 4. ln 2 1 + sin x 1 − sin x 1 = ln 2 (1 + sin x)2 (1 − sin x)(1 + sin x) 1 = ln 2 1 + sin x cos x 2 1 = ln 2 1 = ln 2 (1 + sin x)2 1 − sin2 x 1 = ln 2 1 + sin x 1 + sin x 1 = ln = · 2 ln cos x 2 cos x 1 sin x = ln |sec x + tan x| = ln + cos x cos x 20 (1 + sin x)2 cos2 x