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Integration techniques (Supp. Material 8-IT) I Substitution rule. I Completing the square. I Trigonometric identities. I Polynomial division. I Multiplying by 1. Substitution rule Theorem For every differentiable functions f , u : R → R holds, Z Z 0 f (u(x)) u (x) dx = f (y ) dy . Proof: This is the integral form of the chain rule for derivatives. Let F be a primitive of f , that is, F 0 = f . Then 0 F (u) = F 0 (u) u 0 = f (u) u 0 . Integrate the equation above, Z Z d(F (u)) f (u(x)) u 0 (x) dx = (x) dx = F (u(x)). dx Denoting y = u(x), we get Z Z Z f (u(x)) u 0 (x) dx = F (y ) = F 0 (y ) dy = f (y ) dy . Substitution rule Example Z Evaluate I = 2 3x e tan(2x ) dx. [1 + cos(4x 2 )] Solution: The argument in the tangent function is has an x 2 , and in the integral appears the factor x dx. We try the substitution u = tan(2x 2 ), du = 1 (4x) dx. cos2 (2x 2 ) This substitution will simplify the integration if cos2 (2x 2 ) can be related to [1 + cos(4x 2 )]. And this is the case, since 1 1 + cos(2θ) , θ = 2x 2 , 2 Z Z Z 2 2 3x e tan(2x ) 3x e tan(2x ) 3 u du I = dx = dx = e . [1 + cos(4x 2 )] 2 cos2 (2x 2 ) 2 4 cos2 (θ) = Substitution rule Example Z Evaluate I = 2 3x e tan(2x ) dx. [1 + cos(4x 2 )] 3 Solution: Recall: I = 2 Z 3 I = 8 eu Z du , with u = tan(2x 2 ). 4 e u du = 3 u e + c. 8 We now substitute back with u = tan(2x 2 ), I = 3 tan(2x 2 ) e + c. 8 C Integration techniques (Supp. Material 8-IT) I Substitution rule. I Completing the square. I Trigonometric identities. I Polynomial division. I Multiplying by 1. Completing the square Example Z Evaluate I = dx √ . (x + 3) x 2 + 6x + 4 Solution: The idea is to rewrite the function inside the square root: 6 2 2 x + 4 = x 2 + 2(3x) + 4 x + 6x + 4 = x + 2 2 x 2 + 6x + 4 = [x 2 + 2(3x) + 9] − 9 + 4 = (x + 3)2 − 5. Z dx p u = x + 3, du = dx I = (x + 3) (x + 3)2 − 5 Z |u| du 1 √ I = = √ arcsec √ + c. 5 5 u u2 − 5 |x + 3| 1 We obtain I = √ arcsec √ + c. 5 5 C Completing the square Remark: Sometimes completing the square is not needed. Example Z Evaluate I = (x + 3) dx √ . x 2 + 6x + 4 Solution: Since the factor (x + 3) is in the numerator, instead of the denominator, substitution will work: u = x 2 + 6x + 4, Z I = du = (2x + 6) dx = 2(x + 3) dx. 1 du 1 √ = 2 u 2 We conclude that I = p Z u −1/2 du = 1 2u 1/2 + c. 2 x 2 + 6x + 4 + c. Integration techniques (Supp. Material 8-IT) I Substitution rule. I Completing the square. I Trigonometric identities. I Polynomial division. I Multiplying by 1. C Trigonometric identities Example Z Evaluate I = 2 sec(x) + tan(x) dx. Solution: This problem can be solved using trigonometric identities for sine and cosine functions only. 2 h f (x) = sec(x) + tan(x) = sin(x) i2 (1 + sin(x))2 1 + = . cos(x) cos(x) cos2 (x) 2 sin(x) sin2 (x) 1 1 + 2 sin(x) + sin2 (x) = + + . f (x) = cos2 (x) cos2 (x) cos2 (x) cos2 (x) 1 − cos2 (x) sin2 (x) 1 = − 1. = cos2 (x) cos2 (x) cos2 (x) Trigonometric identities Example Z Evaluate I = 2 sec(x) + tan(x) dx. 2 Solution: Recall: f (x) = sec(x) + tan(x) , 1 2 sin(x) sin2 (x) sin2 (x) 1 f (x) = + + and = − 1. cos2 (x) cos2 (x) cos2 (x) cos2 (x) cos2 (x) 2 sin(x) 2 + − 1. cos2 (x) cos2 (x) Z Z Z u = cos(x), 2 dx 2 sin(x) dx + − dx. I = cos2 (x) cos2 (x) du = − sin(x) dx. Z du 2 I = 2 tan(x) − 2 − x + c ⇒ I = 2 tan(x) + − x + c. u2 cos(x) f (x) = Integration techniques (Supp. Material 8-IT) I Substitution rule. I Completing the square. I Trigonometric identities. I Polynomial division. I Multiplying by 1. Polynomial division Example Z Evaluate I = 4x 2 − 7 dx. 2x + 3 Solution: The degree of the polynomial in the numerator is greater or equal the degree of the polynomial in the denominator. In this case it is convenient to do the division: 2x − 3 2x + 3 4x 2 −7 2 − 4x − 6x − 6x − 7 6x + 9 ⇒ 4x 2 − 7 2 = 2x − 3 + . 2x + 3 2x + 3 2 Z I = Z (2x − 3) dx + 2 dx 2x + 3 ⇒ I = x 2 − 3x + ln(2x + 3) + c. Integration techniques (Supp. Material 8-IT) I Substitution rule. I Completing the square. I Trigonometric identities. I Polynomial division. I Multiplying by 1. Multiplying by 1 Example Z dx . 1 + sin(x) Evaluate I = Solution: Z I = Z 1 − sin(x) 1 dx = 1 + sin(x) 1 − sin(x) Z I = 1 − sin(x) dx. 1 − sin2 (x) Z Z 1 − sin(x) dx sin(x) dx = − dx. cos2 (x) cos2 (x) cos2 (x) 1 and u = cos(x) implies du = − sin(x) dx, cos2 (x) Z du 1 1 I = tan(x) + = tan(x) − + c = tan(x) − + c. u2 u cos(x) Since tan0 (x) = We conclude that I = tan(x) − sec(x) + c. C Multiplying by 1 Example Z Evaluate I = sec(x) dx. Solution: This problem can be solved using trigonometric identities for sine and cosine functions only. sin(x) 1 Z Z + cos2 (x) dx 1 cos2 (x) dx I = = sin(x) 1 cos(x) cos(x) + cos2 (x) Z I = Z I = 1 cos2 (x) 1 cos(x) 1 cos(x) 1 cos(x) + + sin(x) cos2 (x) + sin(x) cos(x) sin(x) cos(x) 1 0 sin(x) = , cos(x) cos2 (x) sin(x) 0 1 . = cos(x) cos2 (x) dx cos2 (x) and 0 1 sin(x) dx ⇒ I = ln + + c. sin(x) cos(x) cos(x) + cos(x)