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- Virginia
- George Mason University
- Mathematics
- Mathematics 213
- Kim
- tcu11_08_05.pdf

Sree Ram M.

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586 Chapter 8: Techniques of Integration Trigonometric Substitutions Trigonometric substitutions can be effective in transforming integrals involving and into integrals we can evaluate directly.2x 2 - a 2 2a 2 + x 2 , 2a 2 - x 2 , 8.5 4100 AWL/Thomas_ch08p553-641 8/20/04 10:07 AM Page 586 8.5 Trigonometric Substitutions 587 H9258H9258H9258 a a a x xx H20857a 2 H11002 x 2 x H11005 a tan H9258 x H11005 a sin H9258 x H11005 a sec H9258 H20857x 2 H11002 a 2 H20857a 2 H11001 x 2 H20857a 2 H11001 x 2 H11005 aH20841sec H9258H20841 H20857a 2 H11002 x 2 H11005 aH20841cos H9258H20841 H20857x 2 H11002 a 2 H11005 aH20841tan H9258H20841 FIGURE 8.2 Reference triangles for the three basic substitutions identifying the sides labeled x and a for each substitution. H9258 H9258 H9258 x a x a x a x a H9266 2 H9266 2 H9266 2 H9266 2 – H9266 2 – H9266 H9258 H11005 sec –1 x a H9258 H11005 sin –1 x a H9258 H11005 tan –1 0 01–1 01–1 FIGURE 8.3 The arctangent, arcsine, and arcsecant of xa, graphed as functions of xa.> > Three Basic Substitutions The most common substitutions are and They come from the reference right triangles in Figure 8.2. x 2 - a 2 = a 2 sec 2 u - a 2 = a 2 ssec 2 u - 1d = a 2 tan 2 u. With x = a sec u, a 2 - x 2 = a 2 - a 2 sin 2 u = a 2 s1 - sin 2 ud = a 2 cos 2 u. With x = a sin u, a 2 + x 2 = a 2 + a 2 tan 2 u = a 2 s1 + tan 2 ud = a 2 sec 2 u. With x = a tan u, x = a sec u.x = a tan u, x = a sin u, We want any substitution we use in an integration to be reversible so that we can change back to the original variable afterward. For example, if we want to be able to set after the integration takes place. If we want to be able to set when we’re done, and similarly for As we know from Section 7.7, the functions in these substitutions have inverses only for selected values of (Figure 8.3). For reversibility, To simplify calculations with the substitution we will restrict its use to inte- grals in which This will place in and make We will then have free of absolute values, provided EXAMPLE 1 Using the Substitution Evaluate L dx 24 + x 2 . x = a tan u a 7 0.2x 2 - a 2 = 2a 2 tan 2 u = ƒ a tan u ƒ = a tan u, tan u Ú 0.[0, p>2dux>a Ú 1. x = a sec u, x = a sec u requires u = sec -1 a x a b with d 0 … u 6 p 2 if x a Ú 1, p 2 6 u … p if x a …-1. x = a sin u requires u = sin -1 a x a b with - p 2 … u … p 2 , x = a tan u requires u = tan -1 a x a b with - p 2 6 u 6 p 2 , u x = a sec u.u = sin -1 sx>ad x = a sin u,u = tan -1 sx>ad x = a tan u, 4100 AWL/Thomas_ch08p553-641 8/20/04 10:07 AM Page 587 Solution We set Then Notice how we expressed in terms of x: We drew a reference triangle for the original substitution (Figure 8.4) and read the ratios from the triangle. EXAMPLE 2 Using the Substitution Evaluate Solution We set Then = 9 2 sin -1 x 3 - x 2 29 - x 2 + C. = 9 2 asin -1 x 3 - x 3 # 29 - x 2 3 b + C = 9 2 su - sin u cos ud + C = 9 2 au - sin 2u 2 b + C = 9 L 1 - cos 2u 2 du = 9 L sin 2 u du L x 2 dx 29 - x 2 = L 9 sin 2 u # 3 cos u du ƒ 3 cos u ƒ 9 - x 2 = 9 - 9 sin 2 u = 9s1 - sin 2 ud = 9 cos 2 u. x = 3 sin u, dx = 3 cos u du, - p 2 6 u 6 p 2 L x 2 dx 29 - x 2 . x = a sin u x = 2 tan u ln ƒ sec u + tan u ƒ = ln ƒ 24 + x 2 + x ƒ + C¿ . = ln ` 24 + x 2 2 + x 2 ` + C = ln ƒ sec u + tan u ƒ + C = L sec u du L dx 24 + x 2 = L 2 sec 2 u du 24 sec 2 u = L sec 2 u du ƒ sec u ƒ 4 + x 2 = 4 + 4 tan 2 u = 4s1 + tan 2 ud = 4 sec 2 u. x = 2 tan u, dx = 2 sec 2 u du, - p 2 6 u 6 p 2 , 588 Chapter 8: Techniques of Integration 2sec 2 u = ƒ sec u ƒ sec u 7 0 for - p 2 6 u 6 p 2 From Fig. 8.4 Taking C¿=C - ln 2 H9258 2 x H208574 H11001 x 2 FIGURE 8.4 Reference triangle for (Example 1): and sec u = 24 + x 2 2 . tan u = x 2 x = 2 tan u cos u 7 0 for - p 2 6 u 6 p 2 sin 2u = 2 sin u cos u Fig. 8.5 H9258 3 x H208579 H11002 x 2 FIGURE 8.5 Reference triangle for (Example 2): and cos u = 29 - x 2 3 . sin u = x 3 x = 3 sin u 4100 AWL/Thomas_ch08p553-641 8/20/04 10:08 AM Page 588 EXAMPLE 3 Using the Substitution Evaluate Solution We first rewrite the radical as to put the radicand in the form We then substitute With these substitutions, we have A trigonometric substitution can sometimes help us to evaluate an integral containing an integer power of a quadratic binomial, as in the next example. EXAMPLE 4 Finding the Volume of a Solid of Revolution Find the volume of the solid generated by revolving about the x-axis the region bounded by the curve the x-axis, and the lines and Solution We sketch the region (Figure 8.7) and use the disk method: To evaluate the integral, we set x 2 + 4 = 4 tan 2 u + 4 = 4stan 2 u + 1d = 4 sec 2 u x = 2 tan u, dx = 2 sec 2 u du, u = tan -1 x 2 , V = L 2 0 p[Rsxd] 2 dx = 16p L 2 0 dx sx 2 + 4d 2 . x = 2.x = 0y = 4>sx 2 + 4d, = 1 5 ln ` 5x 2 + 225x 2 - 4 2 ` + C. = 1 5 L sec u du = 1 5 ln ƒ sec u + tan u ƒ + C L dx 225x 2 - 4 = L dx 52x 2 - s4>25d = L s2>5d sec u tan u du 5 # s2>5d tan u C x 2 - a 2 5 b 2 = 2 5 ƒ tan u ƒ = 2 5 tan u. = 4 25 ssec 2 u - 1d = 4 25 tan 2 u x 2 - a 2 5 b 2 = 4 25 sec 2 u - 4 25 x = 2 5 sec u, dx = 2 5 sec u tan u du, 0 6 u 6 p 2 x 2 - a 2 . = 5 C x 2 - a 2 5 b 2 225x 2 - 4 = B 25ax 2 - 4 25 b L dx 225x 2 - 4 , x 7 2 5 . x = a sec u 8.5 Trigonometric Substitutions 589 0 6 u 6 p>2 tan u 7 0 for Fig. 8.6 H9258 2 5x H2085725x 2 H11002 4 FIGURE 8.6 If then and we can read the values of the other trigonometric functions of from this right triangle (Example 3). u u = sec -1 s5x>2d,0 6 u 6 p>2, x = s2>5dsec u, Rsxd = 4 x 2 + 4 4100 AWL/Thomas_ch08p553-641 8/20/04 10:08 AM Page 589 (Figure 8.8). With these substitutions, EXAMPLE 5 Finding the Area of an Ellipse Find the area enclosed by the ellipse Solution Because the ellipse is symmetric with respect to both axes, the total area A is four times the area in the first quadrant (Figure 8.9). Solving the equation of the ellipse for we get or y = b a 2a 2 - x 2 0 … x … a y 2 b 2 = 1 - x 2 a 2 = a 2 - x 2 a 2 , y Ú 0, x 2 a 2 + y 2 b 2 = 1 = pc p 4 + 1 2 d L 4.04. = p L p>4 0 s1 + cos 2ud du = pcu + sin 2u 2 d 0 p>4 = 16p L p>4 0 2 sec 2 u du 16 sec 4 u = p L p>4 0 2 cos 2 u du = 16p L p>4 0 2 sec 2 u du s4 sec 2 ud 2 V = 16p L 2 0 dx sx 2 + 4d 2 590 Chapter 8: Techniques of Integration x y 02 1 y H11005 4 x 2 H11001 4 (a) 2 0 y H11005 4 x 2 H11001 4 (b) x y FIGURE 8.7 The region (a) and solid (b) in Example 4. H9258 2 x H20857x 2 H11001 4 FIGURE 8.8 Reference triangle for (Example 4).x = 2 tan u 2 cos 2 u = 1 + cos 2u x y 0 a–a –b b FIGURE 8.9 The ellipse in Example 5. x 2 a 2 + y 2 b 2 = 1 u = p>4 when x = 2 u = 0 when x = 0; 4100 AWL/Thomas_ch08p553-641 9/2/04 10:37 AM Page 590 The area of the ellipse is If we get that the area of a circle with radius r is pr 2 .a = b = r = 2abc p 2 + 0 - 0d = pab. = 2abcu + sin 2u 2 d 0 p>2 = 4ab L p>2 0 1 + cos 2u 2 du = 4ab L p>2 0 cos 2 u du = 4 b a L p>2 0 a cos u # a cos u du A = 4 L a 0 b a 2a 2 - x 2 dx 8.5 Trigonometric Substitutions 591 u = p>2 when x = a u = 0 when x = 0; x = a sin u, dx = a cos u du, 4100 AWL/Thomas_ch08p553-641 8/20/04 10:08 AM Page 591 Commercial_CD 4100 AWL/Thomas_ch08p553-641

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