Stewart complement 3c3 trigonometsubstitu stu

Trigonometric Substitution In finding the area of a circle or an ellipse, an integral of the form x sa Ϫ x dx arises, where a Ͼ If it were x xsa Ϫ x dx, the substitution u ෇ a Ϫ x would be effective but, as it stands, x sa Ϫ x dx is more difficult If we change the variable from x to ␪ by the substitution x ෇ a sin ␪, then the identity Ϫ sin 2␪ ෇ cos 2␪ allows us to get rid of the root sign because Խ sa Ϫ x ෇ sa Ϫ a sin 2␪ ෇ sa 2͑1 Ϫ sin 2␪ ͒ ෇ sa cos 2␪ ෇ a cos ␪ Խ Notice the difference between the substitution u ෇ a Ϫ x (in which the new variable is a function of the old one) and the substitution x ෇ a sin ␪ (the old variable is a function of the new one) In general we can make a substitution of the form x ෇ t͑t͒ by using the Substitution Rule in reverse To make our calculations simpler, we assume that t has an inverse function; that is, t is one-to-one In this case, if we replace u by x and x by t in the Substitution Rule (Equation 5.5.4), we obtain y f ͑x͒ dx ෇ y f ͑t͑t͒͒tЈ͑t͒ dt This kind of substitution is called inverse substitution We can make the inverse substitution x ෇ a sin ␪ provided that it defines a one-to-one function This can be accomplished by restricting ␪ to lie in the interval ͓Ϫ␲͞2, ␲͞2͔ In the following table we list trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities In each case the restriction on ␪ is imposed to ensure that the function that defines the substitution is one-to-one (These are the same intervals used in Appendix D in defining the inverse functions.) Table of Trigonometric Substitutions Expression Substitution Identity sa Ϫ x x ෇ a sin ␪, Ϫ ␲ ␲ ഛ␪ഛ 2 Ϫ sin 2␪ ෇ cos 2␪ sa ϩ x x ෇ a tan ␪, Ϫ ␲ ␲ Ͻ␪Ͻ 2 ϩ tan 2␪ ෇ sec 2␪ sx Ϫ a x ෇ a sec ␪, 0ഛ␪Ͻ EXAMPLE Evaluate y ␲ 3␲ or ␲ ഛ ␪ Ͻ 2 sec 2␪ Ϫ ෇ tan 2␪ s9 Ϫ x dx x2 SOLUTION Let x ෇ sin ␪, where Ϫ␲͞2 ഛ ␪ ഛ ␲͞2 Then dx ෇ cos ␪ d␪ and Խ Խ s9 Ϫ x ෇ s9 Ϫ sin 2␪ ෇ s9 cos 2␪ ෇ cos ␪ ෇ cos ␪ (Note that cos ␪ ജ because Ϫ␲͞2 ഛ ␪ ഛ ␲͞2.) Thus, the Inverse Substitution Rule gives cos ␪ s9 Ϫ x y x dx ෇ y sin 2␪ cos ␪ d␪ ෇y cos 2␪ d␪ ෇ y cot 2␪ d␪ sin 2␪ ෇ y ͑csc 2␪ Ϫ 1͒ d␪ ෇ Ϫcot ␪ Ϫ ␪ ϩ C ■ TRIGONOMETRIC SUBSTITUTION Since this is an indefinite integral, we must return to the original variable x This can be done either by using trigonometric identities to express cot ␪ in terms of sin ␪ ෇ x͞3 or by drawing a diagram, as in Figure 1, where ␪ is interpreted as an angle of a right triangle Since sin ␪ ෇ x͞3, we label the opposite side and the hypotenuse as having lengths x and Then the Pythagorean Theorem gives the length of the adjacent side as s9 Ϫ x 2, so we can simply read the value of cot from the figure: x ă 9- FIGURE sin ă= cot x s9 Ϫ x x (Although ␪ Ͼ in the diagram, this expression for cot ␪ is valid even when ␪ Ͻ 0.) Since sin ␪ ෇ x͞3, we have ␪ ෇ sinϪ1͑x͞3͒ and so y ͩͪ x s9 Ϫ x s9 Ϫ x dx ෇ Ϫ Ϫ sinϪ1 x x ϩC EXAMPLE Find the area enclosed by the ellipse x2 y2 ϩ ෇1 a2 b2 y SOLUTION Solving the equation of the ellipse for y, we get (0, b) y2 x2 a2 Ϫ x2 ෇ Ϫ ෇ b2 a2 a2 (a, 0) x y෇Ϯ or b sa Ϫ x a FIGURE Because the ellipse is symmetric with respect to both axes, the total area A is four times the area in the first quadrant (see Figure 2) The part of the ellipse in the first quadrant is given by the function b y ෇ sa Ϫ x 0ഛxഛa a ¥ ≈ + =1 b@ a@ and so A෇y a b sa Ϫ x dx a To evaluate this integral we substitute x ෇ a sin ␪ Then dx ෇ a cos ␪ d␪ To change the limits of integration we note that when x ෇ 0, sin ␪ ෇ 0, so ␪ ෇ 0; when x ෇ a, sin ␪ ෇ 1, so ␪ ෇ ␲͞2 Also Խ Խ sa Ϫ x ෇ sa Ϫ a sin 2␪ ෇ sa cos 2␪ ෇ a cos ␪ ෇ a cos ␪ since ഛ ␪ ഛ ␲͞2 Therefore A෇4 b a y a ෇ 4ab y ␲͞2 [ sa Ϫ x dx ෇ b a cos 2␪ d␪ ෇ 4ab y ෇ 2ab ␪ ϩ 12 sin 2␪ y ␲͞2 ␲͞2 ] ␲͞2 ͩ ෇ 2ab a cos ␪ ؒ a cos ␪ d␪ ͑1 ϩ cos 2␪ ͒ d␪ ͪ ␲ ϩ0Ϫ0 ෇ ␲ ab We have shown that the area of an ellipse with semiaxes a and b is ␲ ab In particular, taking a ෇ b ෇ r, we have proved the famous formula that the area of a circle with radius r is ␲ r NOTE Since the integral in Example was a definite integral, we changed the limits of integration and did not have to convert back to the original variable x ■ TRIGONOMETRIC SUBSTITUTION ■ EXAMPLE Find y x sx 2 ϩ4 dx SOLUTION Let x ෇ tan ␪, Ϫ␲͞2 Ͻ ␪ Ͻ ␲͞2 Then dx ෇ sec 2␪ d␪ and Խ Խ sx ϩ ෇ s4͑tan 2␪ ϩ 1͒ ෇ s4 sec 2␪ ෇ sec ␪ ෇ sec ␪ Thus, we have dx y x sx ෇y ϩ4 2 sec 2␪ d␪ ෇ tan ␪ ؒ sec ␪ sec ␪ y tan ␪ d␪ To evaluate this trigonometric integral we put everything in terms of sin ␪ and cos ␪ : sec␪ cos 2␪ cos ␪ ෇ ؒ ෇ 2 tan ␪ cos ␪ sin ␪ sin 2␪ Therefore, making the substitution u ෇ sin ␪, we have y dx ෇ x sx ϩ 4 ෇ +4 x ă FIGUR E tan ă= cos d sin ␪ y ͩ ͪ Ϫ u y ϩC෇Ϫ du u2 ϩC sin ␪ csc ␪ ϩC We use Figure to determine that csc ␪ ෇ sx ϩ 4͞x and so x dx y x sx EXAMPLE Find y sx ϩ4 ෇Ϫ sx ϩ ϩC 4x x dx ϩ4 SOLUTION It would be possible to use the trigonometric substitution x ෇ tan ␪ here (as in Example 3) But the direct substitution u ෇ x ϩ is simpler, because then du ෇ 2x dx and y NOTE x dx ෇ sx ϩ du y su ෇ su ϩ C ෇ sx ϩ ϩ C Example illustrates the fact that even when trigonometric substitutions are possible, they may not give the easiest solution You should look for a simpler method first ■ EXAMPLE Evaluate y sx dx , where a Ͼ Ϫ a2 SOLUTION We let x ෇ a sec ␪, where Ͻ dx ෇ a sec ␪ tan ␪ d␪ and ␪ Ͻ ␲͞2 or ␲ Ͻ ␪ Ͻ 3␲͞2 Then Խ Խ sx Ϫ a ෇ sa 2͑sec 2␪ Ϫ 1͒ ෇ sa tan 2␪ ෇ a tan ␪ ෇ a tan ␪ Therefore y sx dx a sec ␪ tan ␪ ෇y d␪ Ϫa a tan ␪ Խ Խ ෇ y sec ␪ d␪ ෇ ln sec ␪ ϩ tan ␪ ϩ C ■ TRIGONOMETRIC SUBSTITUTION The triangle in Figure gives tan ␪ ෇ sx Ϫ a 2͞a, so we have x ≈-a@ y sx ă sec ă= x a Խ ෇ ln x ϩ sx Ϫ a Ϫ ln a ϩ C a FIGU RE Ϳ dx x sx Ϫ a ෇ ln ϩ ϩC Ϫa a a Writing C1 ෇ C Ϫ ln a, we have y sx EXAMPLE Find y s3͞2 dx ෇ ln x ϩ sx Ϫ a ϩ C1 Ϫ a2 Խ Խ x3 dx ͑4x ϩ 9͒3͞2 SOLUTION First we note that ͑4x ϩ 9͒3͞2 ෇ ͑s4x ϩ )3 so trigonometric substitution is appropriate Although s4x ϩ is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitu3 tion u ෇ 2x When we combine this with the tangent substitution, we have x ෇ tan ␪, which gives dx ෇ sec ␪ d␪ and s4x ϩ ෇ s9 tan 2␪ ϩ ෇ sec ␪ When x ෇ 0, tan ␪ ෇ 0, so ␪ ෇ 0; when x ෇ 3s3͞2, tan ␪ ෇ s3, so ␪ ෇ ␲͞3 y s3͞2 27 x3 ␲͞3 tan ␪ dx ෇ y 3͞2 ͑4x ϩ 9͒ 27 sec3␪ ෇ 163 y sec 2␪ d␪ tan 3␪ ␲͞3 sin ␪ d␪ ෇ 163 y d␪ sec ␪ cos2␪ ␲͞3 ෇ 163 y Ϫ cos 2␪ sin ␪ d␪ cos 2␪ ␲͞3 Now we substitute u ෇ cos ␪ so that du ෇ Ϫsin ␪ d␪ When ␪ ෇ 0, u ෇ 1; when ␪ ෇ ␲͞3, u ෇ 12 Therefore y s3͞2 1͞2 Ϫ u 1͞2 x3 3 dx ෇ Ϫ du ෇ 16 y ͑1 Ϫ u Ϫ2 ͒ du y 16 3͞2 1 ͑4x ϩ 9͒ u ͫ ͬ ෇ 163 u ϩ EXAMPLE Evaluate x y s3 Ϫ 2x Ϫ x u 1͞2 ෇ 163 [( 12 ϩ 2) Ϫ ͑1 ϩ 1͒] ෇ 323 dx SOLUTION We can transform the integrand into a function for which trigonometric substi- tution is appropriate by first completing the square under the root sign: Ϫ 2x Ϫ x ෇ Ϫ ͑x ϩ 2x͒ ෇ ϩ Ϫ ͑x ϩ 2x ϩ 1͒ ෇ Ϫ ͑x ϩ 1͒2 This suggests that we make the substitution u ෇ x ϩ Then du ෇ dx and x ෇ u Ϫ 1, so x y s3 Ϫ 2x Ϫ x dx ෇ y uϪ1 du s4 Ϫ u TRIGONOMETRIC SUBSTITUTION ■ We now substitute u ෇ sin ␪, giving du ෇ cos ␪ d␪ and s4 Ϫ u ෇ cos ␪, so x y s3 Ϫ 2x Ϫ x 2 sin ␪ Ϫ cos ␪ d␪ cos ␪ dx ෇ y ෇ y ͑2 sin ␪ Ϫ 1͒ d␪ ෇ Ϫ2 cos ␪ Ϫ ␪ ϩ C ͩͪ ෇ Ϫs4 Ϫ u Ϫ sinϪ1 u ϩC ͩ ͪ ෇ Ϫs3 Ϫ 2x Ϫ x Ϫ sinϪ1 xϩ1 ϩC Exercises A Click here for answers Click here for solutions S y 21 y 23 y s5 ϩ 4x Ϫ x 25 y s9x 27 y ͑x 29 y x s1 Ϫ x 1–3 Evaluate the integral using the indicated trigonometric substitution Sketch and label the associated right triangle yx yx dx ; sx Ϫ x ෇ sec ␪ s9 Ϫ x dx ; x ෇ sin ␪ y sx ■ x3 dx ; ϩ9 ■ ■ 4–30 x ෇ tan ␪ ■ ■ ■ ■ ■ ■ ■ ■ Evaluate the integral y y s3 s2 x dx s16 Ϫ x dt t st Ϫ 1 dx x s25 Ϫ x y y sx dx ϩ 16 11 y s1 Ϫ 4x 13 y x sx ϩ dx y y st 2 t5 dt ϩ2 dx dx ϩ 6x Ϫ dx ϩ 2x ϩ 2͒2 ■ ■ y sx Ϫ dx x3 14 y u s5 Ϫ u 15 y x2 dx ͑a Ϫ x ͒3͞2 16 y 17 y sx x dx Ϫ7 18 y ͓͑ax͒ dx ■ ■ y sx 22 y 24 y st 26 y s4x Ϫ x 28 y ͑5 Ϫ 4x Ϫ x 30 y dt sx ϩ dx dt Ϫ 6t ϩ 13 x2 dx dx ■ ■ cos t dt s1 ϩ sin t ␲ ͞2 ■ ■ ■ dx Ϫ b ͔ 3͞2 ■ dx ෇ ln ( x ϩ sx ϩ a ) ϩ C ϩ a2 y sx ͩͪ dx x ෇ sinhϪ1 ϩ a2 a ϩC dx x s16x Ϫ ■ These formulas are connected by Formula 3.9.3 du ͒ 5͞2 (b) Use the hyperbolic substitution x ෇ a sinh t to show that 2 y s25 Ϫ t 31 (a) Use trigonometric substitution to show that x sx ϩ dx y 2 12 x 3s4 Ϫ 9x dx sx Ϫ a dx x4 dx 2͞3 t 20 ■ ■ 10 ■ s1 ϩ x dx x 19 32 Evaluate y ͑x x2 dx ϩ a ͒3͞2 (a) by trigonometric substitution (b) by the hyperbolic substitution x ෇ a sinh t ■ ■ TRIGONOMETRIC SUBSTITUTION 33 Find the average value of f ͑x͒ ෇ sx Ϫ 1͞x, ഛ x ഛ 34 Find the area of the region bounded by the hyperbola 9x Ϫ 4y ෇ 36 and the line x ෇ where ␭ is the charge density per unit length on the rod and ␧0 is the free space permittivity (see the figure) Evaluate the integral to determine an expression for the electric field E͑P͒ y 35 Prove the formula A ෇ r 2␪ for the area of a sector of a circle with radius r and central angle ␪ [Hint: Assume Ͻ ␪ Ͻ ␲͞2 and place the center of the circle at the origin so it has the equation x ϩ y ෇ r Then A is the sum of the area of the triangle POQ and the area of the region PQR in the figure.] y P (a, b) L x P 39 Find the area of the crescent-shaped region (called a lune) bounded by arcs of circles with radii r and R (See the figure.) ă O Q R x ; 36 Evaluate the integral r R y dx x sx Ϫ Graph the integrand and its indefinite integral on the same screen and check that your answer is reasonable ; 37 Use a graph to approximate the roots of the equation x s4 Ϫ x ෇ Ϫ x Then approximate the area bounded by the curve y ෇ x s4 Ϫ x and the line y ෇ Ϫ x 38 A charged rod of length L produces an electric field at point P͑a, b͒ given by E͑P͒ ෇ y LϪa Ϫa ␭b dx 4␲ ␧0 ͑x ϩ b ͒3͞2 40 A water storage tank has the shape of a cylinder with diameter 10 ft It is mounted so that the circular cross-sections are vertical If the depth of the water is ft, what percentage of the total capacity is being used? 41 A torus is generated by rotating the circle x ϩ ͑ y Ϫ R͒2 ෇ r about the x-axis Find the volume enclosed by the torus TRIGONOMETRIC SUBSTITUTION ■ Answers S Խ Խ 19 ln (s1 ϩ x Ϫ 1)͞x ϩ s1 ϩ x ϩ C Click here for solutions 21 23 sinϪ1͑͑x Ϫ 2͒͞3͒ ϩ ͑ x Ϫ 2͒s5 ϩ 4x Ϫ x ϩ C ln 3x ϩ ϩ s9x ϩ 6x Ϫ ϩ C Խ Խ sx Ϫ 9͑͞9x͒ ϩ C ͑x Ϫ 18͒ sx ϩ ϩ C 25 ␲͞24 ϩ s3͞8 Ϫ Ϫs25 Ϫ x 2͑͞25x͒ ϩ C 27 ͓tanϪ1͑x ϩ 1͒ ϩ ͑x ϩ 1͒͑͞x ϩ 2x ϩ 2͔͒ ϩ C ln (sx ϩ 16 ϩ x) ϩ C 11 sinϪ1͑2x͒ ϩ x s1 Ϫ 4x ϩ C Ϫ1 13 sec ͑x͞3͒ Ϫ sx Ϫ 9͑͞2x 2͒ ϩ C 15 (x͞sa Ϫ x ) Ϫ sinϪ1͑x͞a͒ ϩ C 17 sx Ϫ ϩ C 1 64 1215 29 sinϪ1 x ϩ 14 x s1 Ϫ x ϩ C 33 (s48 Ϫ secϪ1 7) 37 0.81, 2; 2.10 39 r sR Ϫ r ϩ ␲r ͞2 Ϫ R arcsin͑r͞R͒ 2 41 2␲ 2Rr ■ TRIGONOMETRIC SUBSTITUTION Solutions: Trigonometric Substitution Let x = sec θ, where ≤ θ < π or π ≤ θ < 3π Then dx = sec θ tan θ dθ and p p √ p x2 − = sec2 θ − = 9(sec2 θ − 1) = tan2 θ = |tan θ| = tan θ for the relevant values of θ Z √ dx = x2 x2 − Z sec θ tan θ dθ = sec2 θ · tan θ R cos θ dθ = Note that − sec(θ + π) = sec θ, so the figure is sufficient for the case π ≤ θ < Let x = tan θ, where − π2 < θ < p Z x2 + = p tan2 θ + = π q sin θ + C = √ x2 − +C x 3π Then dx = sec2 θ dθ and 9(tan2 θ + 1) = √ sec2 θ = |sec θ| = sec θ for the relevant values of θ Z Z 33 tan3 θ sec2 θ dθ = 33 tan3 θ sec θdθ = 33 tan2 θ tan θ sec θ dθ sec θ R¡ ¢ ¢ 3R ¡ =3 sec2 θ − tan θ sec θ dθ = 33 u − du [u = sec θ, du = sec θ tan θ dθ] # " ¡ ¢3/2 √ ¢ ¢ x2 + 3¡ 3¡ 3 x +9 − +C = 3 u − u + C = 3 sec θ − sec θ + C = 3 33 x3 √ dx = x2 + Z = ¡ x2 + ¢3/2 −9 p x2 + + C Let t = sec θ, so dt = sec θ tan θ dθ, t = Z √ Z √ or ⇒ θ= π , ¡ ¢p x2 − 18 x2 + + C and t = ⇒ θ = π Then Z π/3 Z π/3 1 cos2 θ dθ sec θ tan θ dθ = dθ = θ tan θ sec θ π/4 π/4 π/4 R π/3 £ ¤π/3 1 = π/4 (1 + cos 2θ) dθ = θ + sin 2θ π/4 h³ ³ ´ √ √ √ ´ ¢i ¡ π π = 12 π3 + 12 23 − π4 + 12 · = 12 12 + 43 − 12 = 24 + 83 − 14 √ dt = t3 t2 − π/3 sec3 Let x = sin θ, so dx = cos θ dθ Then Z Z 1 √ cos θ dθ dx = 52 sin2 θ · cos θ x2 25 − x2 R 1 csc2 θ dθ = − 25 cot θ + C = 25 √ 25 − x2 =− +C 25 x TRIGONOMETRIC SUBSTITUTION ■ 9 Let x = tan θ, where − π2 < θ < π Then dx = sec2 θ dθ and p √ √ x2 + 16 = 16 tan2 θ + 16 = 16(tan2 θ + 1) √ = 16 sec2 θ = |sec θ| = sec θ for the relevant values of θ Z Z sec2 θ dθ = sec θ dθ = ln |sec θ + tan θ| + C1 sec θ ¯ ¯√ ¯ x2 + 16 ¯√ ¯ x¯ + ¯¯ + C1 = ln ¯ x2 + 16 + x¯ − ln |4| + C1 = ln ¯¯ 4 ¡√ ¢ = ln x2 + 16 + x + C, where C = C1 − ln dx √ = x2 + 16 (Since Z √ x2 + 16 + x > 0, we don’t need the absolute value.) 11 Let 2x = sin θ, where − π2 ≤ θ ≤ π √ dx = 12 cos θ dθ, and − 4x2 = q Then x = sin θ, − (2x)2 = cos θ ¡ ¢ R√ R R − 4x2 dx = cos θ 12 cos θ dθ = 14 (1 + cos 2θ) dθ ¡ ¢ = 14 θ + 12 sin 2θ + C = 14 (θ + sin θ cos θ) + C h i p = 14 sin−1 (2x) + 2x − 4x2 + C 13 Let x = sec θ, where ≤ θ < π2 or π ≤ θ < 3π Then √ dx = sec θ tan θ dθ and x2 − = tan θ, so Z Z Z √ x −9 tan θ tan2 θ dx = sec θ tan θ dθ = dθ 3 x 27 sec θ sec2 θ R sin2 θ dθ = = = ³x´ 1 sec−1 − 6 15 Let x = a sin θ, where − π2 ≤ θ ≤ Z x2 dx (a2 − x2 )3/2 = = Z Z π R (1 − cos 2θ) dθ = 16 θ − sin 2θ + C = 16 θ − sin θ cos θ + C √ ³ x ´ √x2 − x2 − +C + C = sec−1 − x x 2x2 Then dx = a cos θ dθ and a2 sin2 θ a cos θ dθ = a3 cos3 θ ¡ 12 Z tan2 θ dθ ¢ sec2 θ − dθ = tan θ − θ + C x x =√ − sin−1 + C a a2 − x2 17 Let u = x2 − 7, so du = 2x dx Then Z x √ dx = x2 − Z √ du = u ·2 p √ u + C = x2 − + C 10 ■ TRIGONOMETRIC SUBSTITUTION 19 Let x = tan θ, where − π2 < θ < π2 Then dx = sec2 θ dθ √ and + x2 = sec θ, so Z Z √ Z + x2 sec θ sec θ dx = sec2 θ dθ = (1 + tan2 θ) dθ x tan θ tan θ R = (csc θ + sec θ tan θ) dθ = ln |csc θ − cot θ| + sec θ + C [by Exercise 39 in Additional Topics: Trigonometric Integrals] ¯ √ ¯√ ¯ ¯√ ¯ + x2 − ¯ √ ¯ + x2 + x2 ¯¯ ¯ ¯ + + x2 + C ¯ − ¯+ + C = ln ¯ = ln ¯ ¯ x x x 21 Let u = − 9x2 ⇒ R 2/3 du = −18x dx Then x2 = (4 − u) and √ ¡ 1¢ R0 du = x3 − 9x2 dx = 91 (4 − u)u1/2 − 18 = 162 Or: Let 3x = sin θ, where − π2 ≤ θ ≤ h 3/2 3u − 25 u5/2 i4 = 162 162 £ 64 R4³ − 64 4u1/2 u3/2 du Ô = 64 1215 23 + 4x − x2 = −(x2 − 4x + 4) + = −(x − 2)2 + Let x − = sin θ, − π2 ≤ θ ≤ π2 , so dx = cos θ dθ Then R√ Rp Rp + 4x − x2 dx = − (x − 2)2 dx = − sin2 θ cos θ dθ √ R R = cos2 θ cos θ dθ = cos2 θ dθ R ¡ ¢ = 92 (1 + cos 2θ) dθ = 92 θ + 12 sin 2θ + C = 92 θ + = = 9 sin 2θ + C = 92 θ + 94 (2 sin θ cos ) + C ả + 4x − x2 x−2 −1 x − sin + à à +C 3 ả x−2 sin−1 + (x − 2) + 4x − x2 + C let u = sec θ, where ≤ θ < Z du √3 = u2 − Z = π Z Z du dx √ √3 = Now 9x + 6x − u2 − √ or π ≤ θ < 3π Then du = sec θ tan θ dθ and u2 − = tan θ, so 25 9x2 + 6x − = (3x + 1)2 − 9, so let u = 3x + 1, du = 3dx Then √ ¯ ¯ u + u2 − R sec θ tan θ dθ = 13 sec θdθ = 13 ln|sec θ + tan θ| + C1 = 13 ln¯¯ tan θ ¯ ¯ ¯ ¯ p p ¯ ¯ ¯ ¯ ln¯u + u2 − 9¯ + C = 13 ln¯3x + + 9x2 + 6x − ¯ + C ¯ ¯ ¯ + C1 ¯ 27 x2 + 2x + = (x + 1)2 + Let u = x + 1, du = dx Then Z dx = (x2 + 2x + 2)2 Z R du = (u2 + 1)2 Z sec2 θdθ sec4 θ " where u = tan θ, du = sec2 θ dθ, and u2 + = sec2 θ # R cos2 θ dθ = 12 (1 + cos 2θ) dθ = 12 (θ + sin θ cos θ) + C ¸ · · ¸ 1 u x+1 −1 −1 +C = tan u + tan (x + 1) + +C = + u2 x + 2x + = TRIGONOMETRIC SUBSTITUTION ■ 11 29 Let u = x2 , du = 2x dx Then R √ ¡ R√ ¢ x − x4 dx = − u2 12 du = R (1 R cos θ · cos θ dθ " where u = sin θ, du = cos θ dθ, √ and − u2 = cos θ # + cos 2θ)dθ = 14 θ + 18 sin 2θ + C = 14 θ + 14 sin θ cos θ + C √ √ = 14 sin−1 u + 14 u − u2 + C = 14 sin−1 (x2 ) + 14 x2 − x4 + C = 2 31 (a) Let x = a tan θ, where − π2 < θ < Z π Then √ x2 + a2 = a sec θ and ¯ ¯√ Z ¯ x + a2 x ¯¯ a sec2 θ dθ ¯ = sec θ dθ = ln|sec θ + tan θ| + C1 = ln¯ + ¯ + C1 a sec θ a a ³ ´ p = ln x + x2 + a2 + C where C = C1 − ln |a| dx √ = x2 + a2 Z √ (b) Let x = a sinh t, so that dx = a cosh t dt and x2 + a2 = a cosh t Then Z Z dx x a cosh t dt √ = = t + C = sinh−1 + C a cosh t a x2 + a2 √ 33 The average value of f (x) = x2 − 1/x on the interval [1, 7] is " # Z 7√ Z where x = sec θ, dx = sec θ tan θ dθ, x −1 α tan θ dx = · sec θ tan θ dθ √ 7−1 x sec θ x2 − = tan θ, and α = sec−1 R R α α = 16 tan2 θ dθ = 16 (sec2 θ − 1) dθ h iα = 16 tan θ − θ = 16 (tan α − α) ¡√ ¢ = 16 48 − sec−1 35 Area of 4P OQ = 12 (r cos θ)(r sin θ) = 12 r sin θ cos θ Area of region P QR = Let x = r cos u ⇒ dx = −r sin u du for θ ≤ u ≤ π2 Then we obtain Rr r cos θ √ r − x2 dx R√ R R r2 − x2 dx = r sin u (−r sin u) du = −r sin2 u du = − 12 r (u − sin u cos u) + C p = − 12 r cos−1 (x/r) + 12 x r2 − x2 + C so area of region P QR = = h £ −r cos−1 (x/r) + x ¡ p r − x2 − −r θ + r cos θ r sin θ = 12 r θ 12 r sin cos ÂÔ ir r cos θ and thus, (area of sector P OR) = (area of 4P OQ) + (area of region P QR) = 12 r θ 12 ■ TRIGONOMETRIC SUBSTITUTION √ 37 From the graph, it appears that the curve y = x2 − x2 and the line √ y = − x intersect at about x = 0.81 and x = 2, with x2 − x2 > − x on (0.81, 2) So the area bounded by the curve and the line is A ≈ R Ê Ô2 Ô R2 Ê x − x2 − (2 − x) dx = 0.81 x2 − x2 dx − 2x − 12 x2 0.81 0.81 To evaluate the integral, we put x = sin θ, where − π2 ≤ θ ≤ π Then dx = cos θ dθ, x = ⇒ θ = sin−1 = π2 , and x = 0.81 ⇒ θ = sin−1 0.405 ≈ 0.417 So √ R2 R π/2 R π/2 R π/2 x2 − x2 dx ≈ 0.417 sin2 θ (2 cos θ)(2 cos θ dθ) = 0.417 sin2 2θ dθ = 0.417 12 (1 − cos 4) d 0.81 Ê Ô/2 ÊĂ Â Ă ÂÔ = θ − 14 sin 4θ 0.417 = π2 − − 0.417 − 14 (0.995) ≈ 2.81 £¡  à ÂÔ Thus, A 2.81 à − 12 · 22 − · 0.81 − 12 · 0.812 ≈ 2.10 39 Let the equation of the large circle be x2 + y = R2 Then the equation of the small circle is x2 + (y − b)2 = r , √ where b = R2 − r2 is the distance between the centers of the circles The desired area is √ √ R r ÊĂ Â Ô RrĂ Â A = −r b + r2 − x2 − R2 − x2 dx = b + r − x2 − R2 − x2 dx =2 Rr b dx + Rr√ Rr√ r − x2 dx − R2 − x2 dx √ The first integral is just 2br = 2r R2 − r To evaluate the other two integrals, note that R R R√ a2 − x2 dx = a2 cos2 θ dθ [x = a sin θ, dx = a cos θ dθ] = 12 a2 (1 + cos 2θ) dθ ¡ ¢ = 12 a2 θ + 12 sin 2θ + C = 12 a2 (θ + sin θ cos θ) + C ³ x ´ a2 ³ x ´ √a2 − x2 ³x´ xp a2 a2 arcsin + +C = arcsin + a2 − x2 + C = a a a a so the desired area is h ir h ir p p p A = 2r R2 − r + r arcsin(x/r) + x r2 − x2 − R2 arcsin(x/R) + x R2 − x2 = 2r p R2 − r + r 2¡ π ¢ h − R2 arcsin(r/R) + r p i R2 − r = r 41 We use cylindrical shells and assume that R > r x2 = r − (y − R)2 p g(y) = r − (y − R)2 and V = R R+r R−r 2πy · Rr p r2 − (y − R)2 dy = Rr −r p R2 − r + ⇒ x=± π 2r p − R2 arcsin(r/R) r2 − (y − R)2 , so √ 4π(u + R) r2 − u2 du [where u = y − R] √ Rr √ = 4π −r u r2 − u2 du + 4πR −r r − u2 du " where u = r sin θ, du = r cos θ dθ in the second integral # h ¡ ¢3/2 ir R R π/2 π/2 = 4π − 13 r − u2 + 4πR −π/2 r2 cos2 θ dθ = − 4π cos2 θ dθ (0 − 0) + 4πRr −π/2 = 2πRr R π/2 −π/2 −r £ (1 + cos 2θ) dθ = 2πRr θ + sin 2θ Another method: Use washers instead of shells, so V = 8πR the integral using y = r sin Ô/2 /2 Rrp = 22 Rr2 r − y2 dy as in Exercise 6.2.39(a), but evaluate