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Calculus II Lecture 7 The freecalc project team Reference Lectures https://sourceforge.net/projects/freecalculus/ 2016 The freecalc team Lecture 7 2016 Outline √ 1 Integrals of form R(x, ax 2√+ bx + c)dx, function √ R - rational √ Transforming to the forms x 2 + 1, −x 2 + 1, x 2 − 1 Table of Euler √ and trig substitutions The case √x 2 + 1 The case √−x 2 + 1 The case x 2 − 1 2 Rationalizing Substitutions R The freecalc team Lecture 7 2016 License to use and redistribute These lecture slides and their LATEX source code are licensed to you under the Creative Commons license CC BY 3.0. You are free to Share - to copy, distribute and transmit the work, to Remix - to adapt, change, etc., the work, to make commercial use of the work, as long as you reasonably acknowledge the original project. Latest version of the .tex sources of the slides: https: //sourceforge.net/p/freecalculus/code/HEAD/tree/ Should the link be outdated/moved, search for “freecalc project”. Creative Commons license CC BY 3.0: https://creativecommons.org/licenses/by/3.0/us/ and the links therein. The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Integrals of form function R √ R(x, ax 2 + bx + c)dx, R - rational Let R(x, y ) be an arbitrary rational expression in two variables (quotient of polynomials in two variables). Question Can we integrate Z p R x, ax 2 + bx + c dx? Yes. We will learn how in what follows. The algorithm for integration is roughly: to one of √ substitution √ RUse linear R to transform R three√integrals: R(x, x 2 + 1)dx, R(x, −x 2 + 1)dx, R(x, x 2 − 1)dx. Use trigonometric substitution or Euler substitution to transform to trigonometric or rational function integral (no radicals). Solve as previously studied. We motivate why we need such integrals by examples such as computing the area of an ellipse. The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Trigonometric Substitution To R √find the area of a circle or ellipse, one needs to compute a2 − x 2 dx. R √ For x a2 − x 2 dx, the substitution u = a2 − x 2 would work. R√ For a2 − x 2 dx, we need a more elaborate substitution. Instead, substitute x = a sin θ. q q p p a2 − x 2 = a2 − a2 sin2 θ = a2 (1 − sin2 θ) = a2 cos2 θ = a| cos θ|. With u = a2 − x 2 , the new variable is a function of the old one. With x = a sin θ, the old variable is a function of the new one. The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Transforming to the forms p x 2 + 1, p −x 2 + 1, p x2 − 1 p Linear substitutions to simplify radicals ay 2 + by + c p Using linear substitutions, radicals of form ay 2 + by + c, a 6= 0, b2 −√ 4ac 6= 0 can be transformed to (multiple of): 2 √x + 1 2 √−x + 1 x 2 − 1. We already studied how to do that using completing the square when dealing with rational functions. The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Transforming to the forms p x 2 + 1, p −x 2 + 1, Recall: linear substitution is subst. of the form u = px + q. Example √ 2 + x + 1 to multiple of Use linear substitution to transform x r √ 1 1 1 2 2 x +x +1 = x +2· x + − +1 2 4 4 s 2 1 3 = + x+ 2 4 v ! u 2 u3 4 1 t = x+ +1 4 3 2 √ s 2 3 2 1 √ = x+ +1 2 3 √2 3p 2 = u + 1, √ √ 2 2 1 2 3 3 √ where u = x+ = x+ . 2 3 3 3 The freecalc team Lecture 7 √ 2016 p x2 − 1 u 2 + 1. Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Transforming to the forms p x 2 + 1, p −x 2 + 1, p x2 − 1 Recall: linear substitution is subst. of the form u = px + q. Example √ 2 + x + 1 to multiple of Use linear subst. to transform −2x q √ 1 1 2 2 − 2 x − 2x − 2 − 2x + x + 1 = q 1 1 1 1 2 = −2 x − 2 4 x + 16 − 16 − 2 r 2 9 = − 2 x − 14 − 16 r 2 9 1 16 = − x − +1 8 9 4 q 2 3 4 1 = √8 − 3 x − 4 +1 √ 3 = √8 −u 2 + 1, 4 1 4 where u = 3 x − 4 = 3 x − 31 . The freecalc team Lecture 7 √ −u 2 + 1. 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Table of Euler and trig substitutions Let R be a rational function in two variables. So far, transformations we converted all integrals of the Z with linear p form R(x, ax 2 + bx + c)dx to one of the three forms: √ √ √ R R R R(x, x 2 + 1)dx, R(x, −x 2 + 1)dx , R(x, x 2 − 1)dx. Each of the above integrals can be transformed to a rational trigonometric integral using 3 pairs of substitutions: x = tan θ, x = cot θ; x = sin θ, x = cos θ; x = csc θ, x = sec θ. We studied that trigonometric integrals are converted to rational function integrals via θ = 2 arctan t. The resulting 3 pairs of substitutions are called Euler substitutions: x = tan(2 arctan t), x = cot(2 arctan t); x = sin(2 arctan t), x = cos(2 arctan t); x = csc(2 arctan t), x = sec(2 arctan t). The Euler substitutions directly transform the integral to a rational function integral. We will demonstrate that the Euler substitutions are rational. The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Table of Euler and trig substitutions Trigonometric substitution and Euler substitution Expression Substitution Variable range Relevant identity π π 2 2θ √ x = tan θ θ ∈ − 1 + tan θ = sec , 2 2 x2 + 1 x = cot θ θ ∈ (0, π) 1 + cot2 θ = csc2 θ π π 2 2θ √ , x = sin θ θ ∈ − 1 − sin θ = cos 2 2 −x 2 + 1 x = cos θ θ ∈ (0, π) 1 − cos2 θ = cos2 θ π 3π 2 θ − 1 = cot2 θ √ x = csc θ θ ∈ 0, ∪ π, csc 2 π2 3π x2 − 1 x = sec θ θ ∈ 0, 2 ∪ π, 2 sec2 θ − 1 = tan2 θ Euler substitution by applying in addition θ = 2 arctan t 2t √ x = −1 < t < 1 (?) 2 2 1−t x +1 x = 12 1t − t 0 < t (?) 2t √ x = −1 ≤ t ≤ 1 (?) 2 2 1+t −x + 1 1−t 2 x = 1+t 2 0<t (?) 1 1 √ x = 2 t + t t ∈ (−∞, −1) ∪ [0, 1) (?) 2 x −1 1+t 2 x = 1−t 2 t ∈ (−∞, −1) ∪ [0, 1) (?) The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 + 1 Trigonometric substitution x = cot θ for √ x2 + 1 √ The trigonometric substitution x = cot θ, θ ∈ (0, π) for x 2 + 1: p p cot 2 θ + 1 x2 + 1 = s cos2 θ = +1 2 s sin θ cos2 θ + sin2 θ = sin2 θ s when θ ∈ (0, π) we have 1 1 sin θ ≥ 0 and so p = = p sin2 θ sin2 θ sin2 θ = sin θ 1 = = csc θ . sin θ The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 + 1 Trigonometric substitution x = cot θ for √ x2 + 1 √ The trigonometric substitution x = cot θ, θ ∈ (0, π) for x 2 + 1: p 1 2 = csc θ . x +1 = sin θ The differential dx can be expressed via dθ from x = cot θ. To summarize: Definition √ The trigonometric substitution x = cot θ, θ ∈ (0, π) for x 2 + 1 is given by: x = cot θ √ 1 2 x +1 = = csc θ sin θ dθ 2 dx = − = − csc θ?dθ 2 sin θ θ = arccotx . The freecalc team Lecture 7 2016 Integrals of form R R(x, Example Z x2 √ p ax 2 + bx + c)dx, R - rational function 1 x2 +9 dx = The freecalc team 1 Z q x 23 x 2 The case dx p x2 + 1 +1 Z 1 √ = d(3 cot θ) 2 2 Z (3 cot θ) 3 cot θ+ 1 1 2 √ = − 3 csc θ dθ 2 2 Z27 cot θ 2 csc θ 1 − csc θ = dθ 2 9 Z cot θ csc θ Z 1 − sin θ 1 1 = dθ = d( cos θ) 2 2 9 9 cos θ cos θ Z 1 du 1 sec θ = = − +C =− +C 2 9 9u 9 u √ x2 + 9 =− +C 9x 3 Lecture 7 Set x 3 = cot θ x = 3 cot θ θ ∈ (0, π) θ ∈ (0, π) ⇒ csc θ > 0 Set u = cos θ p x2 + 9 θ x 2016 3 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Euler subst. for √ The case p x2 + 1 x 2 + 1 corresponding to x = cot θ √ x = cot θ transforms dx, x, x 2 + 1 to trig form. θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form. What if we compose the above? We get the Euler substitution: x = = = = = cot θ cot (2 arctan t) 1 − tan2 (arctan t) 2 tan(arctan t) 1 − t2 2t 1 1 −t . 2 t The freecalc team cos(2z) 1 − tan2 z |Recall: cot(2z) = = sin(2z) 2 tan z Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Euler subst. for √ The case p x2 + 1 x 2 + 1 corresponding to x = cot θ √ x = cot θ transforms dx, x, x 2 + 1 to trig form. θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form. What if we compose the above? We get the Euler substitution: 1 1 x = −t . 2 t We can furthermore compute s 2 p 1 1 2 x +1 = −t +1 4 t s 2 1 1 = −t +4 2 t The freecalc team Lecture 7 1 t −t 2 +4= 1 t +t 2016 2 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Euler subst. for √ The case p x2 + 1 x 2 + 1 corresponding to x = cot θ √ x = cot θ transforms dx, x, x 2 + 1 to trig form. θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form. What if we compose the above? We get the Euler substitution: 1 1 x = −t . 2 t We can furthermore compute s 2 p 1 1 2 x +1 = −t +1 4 t s 2 1 1 = +t 2 t 1 1 = +t . 2 t The freecalc team Lecture 7 q 1 t 2 + t = 1t + t because t > 0 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Euler subst. for √ The case p x2 + 1 x 2 + 1 corresponding to x = cot θ √ x = cot θ transforms dx, x, x 2 + 1 to trig form. θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form. What if we compose the above? We get the Euler substitution: 1 1 x = −t . 2 t We can furthermore compute p 1 1 2 x +1 = +t . 2 t Finally compute dx = t = The freecalc team 1 1 1 1 d −t = − + 1 dt 2 2 t 2 t p 1 1 1 1 +t − − t = x2 + 1 − x 2 t 2 t Lecture 7 . 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Euler subst. for √ The case p x2 + 1 x 2 + 1 corresponding to x = cot θ √ x = cot θ transforms dx, x, x 2 + 1 to trig form. θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form. What if we compose the above? We get the Euler substitution: Definition The Euler substitution for by: The freecalc team √ x 2 + 1 corresponding to x = cot θ is given 1 1 −t , t >0 2 t 1 1 +t 2 t 1 1 − + 1 dt 2 √ 2 t x2 + 1 − x . x = p x2 + 1 = dx = t = Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function 1 2 Euler substitution: x = √ t = x 2 + 1 − x, dx = − 21 Example 1 t The case p x2 + 1 √ 1 1 2 −t , x +1= 2 t +t , 1 + 1 dt. Recall t > 0. t2 Z p Z 1 1 1 1 2 +t + 1 dt x + 1 dx= − 2 2 t Z 2 t 1 1 1 =− + 2 + t dt 3 4 t t 2 −2 t 1 t + 2 ln |t| + = − − +C 2 4 2 1 −1 1 1 1 −1 t −t t +t − ln t + C = 2 2 2 2 p 1 p 2 1 = x x + 1 − ln x2 + 1 − x + C 2 2 √ p 2+1+x x 1 1 √ +C = x x 2 + 1 + ln √ 2 2 x2 + 1 − x x2 + 1 + x 1 p 2 1 p 2 = x x + 1 + ln x +1+x +C 2 2 The freecalc team Lecture 7 2016 Integrals of form R R(x, Example p ax 2 + bx + c)dx, R - rational function The case p x2 + 1 √ √ 2 Find the area locked b-n the hyperbolas y = ± x + 1 and x = ±2 2. y = p x2 + 1 1 2 We studied v = u is called a u why do we call hyperbola: y+x y+x v =0 ( 2 , 2 ) v √ y +x =0 y = x 2 + 1 hyperbola? Compute: (y, x) √ x2 + 1 = y x2 + 1 = y2 y −x =0 2 − x2 u=0 y = 1 √ √ 2 2 1 p (y − x) (y + x) = 2 y =− x +1 2 2 2 1 uv = Signed distance b-n (x, y ) and line 2 1 u =r0 equals v = u2 , 2 2 (x+y) (x+y) √ x− 2 + y− 2 ± 2 u = (y − x) q 2 √ √ where . Consider 2 1 2 2 v = 2 (y + x) = ± 2 (y − x) = ± 2 (y − x) = an arbitrary point (x, y). u. The freecalc team (x, y) Lecture 7 2016 Integrals of form R R(x, Example p ax 2 + bx + c)dx, R - rational function The case p x2 + 1 √ √ 2 Find the area locked b-n the hyperbolas y = ± x + 1 and x = ±2 2. y = p 1 2 x2 + 1 We studied v = u is called a why do we call hyperbola: v =0 √ y +x =0 y = x 2 + 1 hyperbola? Compute: √ x2 + 1 = y x2 + 1 = y2 y −x =0 2 − x2 u=0 y = 1 √ √ 2 2 1 p (y − x) (y + x) = 2 y =− x +1 2 2 2 1 uv = Signed distance b-n (x, y ) and line 2 1 u = 0 equals u. Similarly v = u2 , compute that signed distance b-n √ (x, y) and the line v = 0 equals v . u = √22 (y − x) where . Consider 2 ⇒ y 2 − x 2 = 1 is the hyperbola v = 2 (y + x) 1/2 v = u in the (u, v )-plane. an arbitrary point (x, y). The freecalc team Lecture 7 2016 Integrals of form R R(x, Example p ax 2 + bx + c)dx, R - rational function The case p x2 + 1 √ √ 2 Find the area locked b-n the hyperbolas y = ± x + 1 and x = ±2 2. y = p x2 + 1 The √area in question is: 2 2 Z p 2 x 2 + 1dx √ −2 h2 y =− p x2 + 1 p = 2 x x2 + 1 p i2√2 + ln x2 + 1 + x √ 0−2 2 q √ √ = 2 2 2 (2 2)2 + 1 q √ √ + ln (2 2)2 + 1 + 2 2 √ √ = 12 2 + 2 ln 3 + 2 2 ≈ 20.496 The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 + 1 Example Find R √x x 2 +4 dx. We could use the trig substitution x = 2 tan θ. But there is an easier way: u = x 2 + 4. du =Z2xdx. Z p √ 1 du x √ √ = u + C = x2 + 4 + C dx = 2 u x2 + 4 The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p −x 2 + 1 Trigonometric substitution x = cos θ for √ −x 2 + 1 √ The trigonometric substitution x = cos θ, θ ∈ [0, π] for −x 2 + 1: p p 1 − cos2 θ −x 2 + 1 = p when θ ∈ [0, π] we have 2 p = sin θ sin θ ≥ 0 and so sin2 θ = sin θ = sin θ . To summarize: Definition The trigonometric substitution x = cos θ, θ ∈ [0, π] for given by: x = cos θ √ −x 2 + 1 = sin θ dx = − sin θdθ θ = arccos x . The freecalc team Lecture 7 √ −x 2 + 1 is 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p −x 2 + 1 Example Evaluate R √9−x 2 x2 dx. Let x = 3 sin θ, where −π/2 ≤ θ ≤ π/2. θ Then dx = 3 cos θdθ. q p p 2 9 − x 2 = 9 − 9 sin θ = 9 cos2 θ = 3| cos θ| = 3 cos θ Z √ 9− x2 x2 dx Z = = = The freecalc team 3 cos θ 3 cos θdθ = Z cot2 θdθ 2 9 sin θ Z (csc2 θ − 1)dθ = − cot θ − θ + C √ x 9 − x2 − − arcsin +C x 3 Lecture 7 2016 Integrals of form R R(x, Example p ax 2 + bx + c)dx, R - rational function The case Find the area enclosed by the ellipse y =b (−a, 0) r x2 a2 + p −x 2 + 1 y2 b2 = 1, a, b > 0. 2 1 − x2 a (a, 0) r 2 y = −b 1 − x 2 a The area in question is Za r x2 2b 1 − 2 dx a −a Za r x2 = 4 b 1 − 2 dx. a Express y via x: x2 y2 + 2 = 1 2 a b y2 x2 = 1− 2 2 b a 2 x 2 2 y = b 1− 2 r a x2 y = ±b 1 − 2 a 0 The freecalc team Lecture 7 2016 Integrals of form R R(x, Example p ax 2 + bx + c)dx, R - rational function The case Find the area enclosed by the ellipse y =b r 2 1 − x2 a r 2 y = −b 1 − x 2 a The area in question is Za r x2 2b 1 − 2 dx a −a Za r x2 = 4 b 1 − 2 dx a x2 a2 The freecalc team −x 2 + 1 y2 b2 = 1, a, b > 0. Trig subst.: set x = a sin θ, θ ∈ 0, . q q x2 a2 sin2 θ Compute: 1 − a2 = 1 − a2 = p 1 − sin2 θ = cos θ. When x = 0, θ = 0 and when x = a, θ = π2 . Z ar Z π 2 2 x 1 − 2 dx = cos θ d (a sin θ) a 0 0 Z π π 2 = = = = = 0 = 4b aπ 4 = πab + p . Lecture 7 a 2 Z0 π cos2 θdθ cos(2θ) + 1 a dθ 2 h0 iθ= π 2 sin(2θ) θ a +2 4 θ=0 π a 0 + 4 − (0 + 0) 2 aπ 4 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p −x 2 + 1 Example Evaluate R √ x dx. 2 3−2x−x Complete the square under the root sign: 3 − 2x − x 2 = 3 + 1 − (x 2 + 2x + 1) = 4 − (x + 1)2 Substitute u = x + 1. Then du = dx and x = u − 1. R R R u−1 x x √ dx = √ dx = √ 2 du 2 2 3−2x−x 4−(x+1) 4−u Let u = 2 sin θ, where −π/2 ≤ θ ≤ π/2. Then du = 2 cos θdθ. p √ √ 2 4 − u 2 = 4 − 4 sin θ = 4 cos2 θ = 2| cos θ| = 2 cos θ R R u−1 R 2 sin θ−1 x √ √ dx = du = 2 cos θdθ 2 cos θ 2 2 4−u R 3−2x−x = (2 − 2 cos θ − θ + C √sin θ − 1)dθ = −1 u 2 = − √4 − u − sin 2 +C = − 3 − 2x − x 2 − sin−1 x+1 +C 2 The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Euler subst. for √ The case p −x 2 + 1 −x 2 + 1 corresponding to x = cos θ √ x = cos θ transforms dx, x, −x 2 + 1 to trig form. θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form. What if we compose the above? We get the Euler substitution: x = = = = cos θ cos(2 arctan t) 1 − tan2 ( arctan t) 1 + tan2 ( arctan t) 1 − t2 1 + t2 The freecalc team 1 − tan2 z cos(2z) = 1 + tan2 z Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Euler subst. for √ The case p −x 2 + 1 −x 2 + 1 corresponding to x = cos θ √ x = cos θ transforms dx, x, −x 2 + 1 to trig form. θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form. What if we compose the above? We get the Euler substitution: x √ −x 2 +1 = = = = = The freecalc team 1 − t2 2 1 + t s t2 1− 1 + t2 2 1− s (1 + t 2 )2 − (1 − t 2 )2 (1 + t 2 )2 s 4t 2 (1 + t 2 )2 (1 + t 2 )2 − (1 − t 2 )2 = 4t 2 √ 4t 2 = 2t because t > 0 2t 1 + t2 Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Euler subst. for √ The case p −x 2 + 1 −x 2 + 1 corresponding to x = cos θ √ x = cos θ transforms dx, x, −x 2 + 1 to trig form. θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form. What if we compose the above? We get the Euler substitution: √ x = −x 2 + 1 = t = dx = = The freecalc team 1 − t2 1 + t2 2t 1 + t2 √ √ √ 1−x 1+x −x 2 + 1 √ √ = 1+x 1+x x +1 2 2 1−t 2 − (1 + t ) d =d 2 1 + t2 1 + t 2 4t d −1 =− dt 2 2 2 1+t (1 + t ) Lecture 7 we use t > 0 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function Euler subst. for √ The case p −x 2 + 1 −x 2 + 1 corresponding to x = cos θ √ x = cos θ transforms dx, x, −x 2 + 1 to trig form. θ = 2 arctan t, t > 0 transforms dθ, cos θ, sin θ to rational form. What if we compose the above? We get the Euler substitution: Definition The Euler substitution for given by: √ −x 2 + 1 corresponding to x = cos θ is x √ The freecalc team = −x 2 + 1 = dx = t = 1 − t2 , t >0 2 1+t 2t 1 + t2 4t dt − 2 2 √(t + 1) −x 2 + 1 . x +1 Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 − 1 Trigonometric substitution x = sec θ for √ x2 − 1 3π ∪ π, 2 : The trigonometric substitution x = sec θ, θ ∈ 0, p p x 2 − 1 = rsec2 θ − 1 1 = −1 2 s cos θ sin2 θ = cos2 θ π 3π p when θ ∈ θ ∈ 0, p 2 ∪ π, 2 we have 2 = tan θ tan θ ≥ 0 and so tan2 θ = tan θ = tan θ . The freecalc team Lecture 7 π 2 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 − 1 Trigonometric substitution x = sec θ for The trigonometric substitution x = sec θ, θ ∈ 0, p x 2 − 1 = tan θ . Definition π 2 √ x2 − 1 3π ∪ π, 2 : √ The trigonometric substitution x = sec θ, θ ∈ (0, π) for x 2 + 1 is given by: 1 π 3π x = sec θ = θ ∈ 0, ∪ π, cos θ 2 2 p x 2 − 1 = tan θ sin θ dθ = sec θ tan θdθ dx = 2 cos θ θ = arcsecx . The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 − 1 Example Find R √ dx x 2 −a2 , a > 0. x = a sec θ, 0 < θ < π/2 or π < θ < 3π/2. θ dx = a sec θ tan θdθ. p p p x 2 − a2 = a2 sec2 θ − a2 = a2 tan2 θ = a| tan θ| = a tan θ Z Z Z dx a sec θ tan θdθ √ = = sec θdθ a tan θ x 2 − a2 √ x x 2 − a2 = ln | sec θ + tan θ| + C = ln + +C a a p = ln x + x 2 − a2 + C1 The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 − 1 Euler substitution x = sec θ, θ = 2 arctan t √ x = sec θ transforms dx, x, x 2 − 1 to trig form. θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ. What if we compose the above? We get the Euler substitution: 1 x = sec θ = cos θ 1 1 − tan2 z = cos(2z) = cos(2 arctan t) 1 + tan2 z 1 + tan2 ( arctan t) = 1 − tan2 ( arctan t) 1 + t2 2 − (1 − t 2 ) = = 2 1 − t2 1−t 2 = −1+ 1 − t2 The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 − 1 Euler substitution x = sec θ, θ = 2 arctan t √ x = sec θ transforms dx, x, x 2 − 1 to trig form. θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ. What if we compose the above? We get the Euler substitution: 2 x = −1+ 1 − t2 s 2 2 √ 1+t 2 x −1 = −1 2 s 1−t = s = = The freecalc team (1 + t 2 )2 − (1 − t 2 )2 (1 − t 2 )2 (1 + t 2 )2 − (1 − t 2 )2 = 4t 2 4t 2 (1 − t 2 )2 t, 1 − t 2 have same sign when t ∈ (−∞, −1) ∪ [0, 1) 2t 1 − t2 Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 − 1 Euler substitution x = sec θ, θ = 2 arctan t √ x = sec θ transforms dx, x, x 2 − 1 to trig form. θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ. What if we compose the above? We get the Euler substitution: 2 x = −1+ 1 − t2 √ 2t 2 x −1 = 1 − t2 1 + t2 x = 1 − t2 (1 − t 2 )x = 1 + t 2 (1 + x)t 2 = x − 1 x −1 2 t = x +r1 x −1 t = ± x +1 The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 − 1 Euler substitution x = sec θ, θ = 2 arctan t √ x = sec θ transforms dx, x, x 2 − 1 to trig form. θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ. What if we compose the above? We get the Euler substitution: 2 x = −1+ 1 − t2 √ 2t 2 x −1 = 1 −rt 2 x −1 t = ± x +1 2 dx = d − 1 + 1 − t2 4t = dt 2 2 (1 − t ) The freecalc team Lecture 7 2016 Integrals of form R R(x, p ax 2 + bx + c)dx, R - rational function The case p x2 − 1 Euler substitution x = sec θ, θ = 2 arctan t √ x = sec θ transforms dx, x, x 2 − 1 to trig form. θ = 2 arctan t, t ∈ (−∞, −1) ∪ [0, 1) rationalizes dθ, cos θ, sin θ. What if we compose the above? We get the Euler substitution: Definition The Euler substitution for by: √ x = x2 − 1 = dx = t = The freecalc team √ x 2 − 1 corresponding to x = sec θ is given 1 + t2 , 2 1−t 2t 1 − t2 4t dt 2 2 (1 − √t ) x2 − 1 ± x +1 Lecture 7 t ∈ (−∞, −1) ∪ [0, 1) . 2016 Rationalizing Substitutions Rationalizing Substitutions Some nonrational fractions can be changed into rational fractions by means of appropriate substitutions.p In particular, when an integrand p n contains an expression of the form g(x), the substitution u = n g(x) may be effective. The freecalc team Lecture 7 2016 Rationalizing Substitutions Example √ Let u = x + 4. Then u 2 = x + 4, so x = u 2 − 4 and dx = 2udu. Z √ Z u x +4 2udu dx = 2 x Z u −24 u =2 du 2 u −4 Z 4 =2 du long division 1+ 2 uZ − 4 Z du = 2 du + 8 u2 − 4 ! Z Z 1 1 4 − 4 du partial fractions = 2 du + 8 u−2 u+2 = 2u + 2(ln |u − 2| − ln |u + 2|) √ +C √ x + 4 − 2 +C = 2 x + 4 + 2 ln √ x + 4 + 2 The freecalc team Lecture 7 2016 Rationalizing Substitutions Example Find the surfaceqarea of the ellipsoid obtained y by rotating y = 1− x2 2 about the x axis. x z The freecalc team Lecture 7 2016 Rationalizing Substitutions Example Find the surfaceqarea of the ellipsoid obtained y by rotating y = y0 = x z (y 0 )2 = Area = = = = = The freecalc team 1− x2 2 about the x axis. 1 q(−x) 2 2 1− x2 x2 4y 2 x = − 2y R2 q x2 1 + 4y 2 dx −2 2πy q 2 2 R2 4y +x 2πy dx 2 −2 4y q 2 2 R2 4y +x 2πy dx −2 r 4y 2 R2 2 x 2 dx π 4 1 − + x 2 −2 R2 √ 2 dx 4 − x π −2 Lecture 7 2016