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LESSON 9 INTEGRATION BY TRIGONOMETRIC SUBSTITUTION 2 2 n/2 If the integrand of an integral contains an expression of the form ( a x ) , where a is a positive real number and n is an odd integer, then the domain of the integrand is either the closed interval [ a , a ] or is a subset of this interval. Then try the trigonometric substitution: Let x a sin , where . Then 2 2 dx a cos d . Note that the range of the function x a sin is the closed 2 2 2 2 2 2 2 interval [ a , a ] . Also, we have that a x = a ( a sin ) = a a sin = a 2 (1 sin 2 ) = a 2 cos 2 . Thus, ( a 2 x 2 ) n / 2 = ( a 2 cos 2 ) n / 2 = ( a 2 cos 2 ) n = ( a cos ) n = a cos . Since a 0 , then a a . Since cos 0 whenever , then cos cos . Thus, ( a 2 x 2 ) n / 2 = a n cos n . 2 2 n n If x a sin , then sin x . Using right triangle trigonometry, we have that a a x a2 x2 cos a2 x2 , sec a a a2 x2 , tan x a2 x2 , cot a2 x2 x 2 2 n/2 If the integrand of an integral contains an expression of the form ( a x ) , where a is a positive real number and n is an integer, then the domain of the integrand is the set of all real numbers. Then try the trigonometric substitution: 2 Let x a tan , where . Then dx a sec d . Note that the range 2 2 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 of the function x a tan is the set of all real numbers. Also, we have that a 2 x 2 = a 2 ( a tan ) 2 = a 2 a 2 tan 2 = a 2 (1 tan 2 ) = a 2 sec 2 . Thus, ( a 2 x 2 ) n / 2 = ( a 2 sec 2 ) n / 2 = ( a 2 sec 2 ) n = ( a sec ) n = a sec . Since a 0 , then a a . Since sec 1 0 whenever , then 2 2 2 2 n/2 n n sec sec . Thus, ( a x ) = a sec . n If x a tan , then tan n x . Using right triangle trigonometry, we have that a a2 x2 x a cos a a2 x2 , sec a2 x2 , sin a x a2 x2 , csc a2 x2 x 2 2 n/2 If the integrand of an integral contains an expression of the form ( x a ) , where a is a positive real number and n is an odd integer, then the domain of the integrand is the set ( , a ] [ a , ) or is a subset of this set. Then try the 3 trigonometric substitution: Let x a sec , where 0 or . 2 2 Then dx a sec tan d . Note that the range of the function x a sec is the set ( , a ] [ a , ) . Also, we have that x 2 a 2 = ( a sec ) 2 a 2 = a 2 sec 2 a 2 = a 2 ( sec 2 1) = a 2 tan 2 . Thus, ( x 2 a 2 ) n / 2 = ( a 2 tan 2 ) n / 2 tan . Since a 0 , then a a . Since 3 tan 0 whenever 0 or , then tan tan . Thus, 2 2 ( x 2 a 2 ) n / 2 = a n tan n . 2 2 n n = ( a tan ) = ( a tan ) = a n n Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 If x a sec , then sec x . Using right triangle trigonometry, we have that a x x2 a2 a sin x2 a2 , csc x x x2 a2 , tan x2 a2 , cot a a x2 a2 Examples Evaluate the following integrals. 1. dx x2 9 Note that we do not have the x in the integral to do the simple substitution to 2 let u x 9 . We will use the technique of Trigonometric Substitution. Let x 3 tan , where . Then dx 3sec 2 d . 2 2 2 2 2 2 Also, x 9 9 tan 9 = 9 ( tan 1 ) = 9 sec . x 2 9 9 sec 2 3 sec 3sec since sec 1 0 . Thus, we have that when 2 2 Thus, dx x2 9 = 3 sec 2 d = 3 sec sec d = ln sec tan c1 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 Since x 3 tan , then tan have that sec x . Using right triangle trigonometry, we 3 x2 9 . 3 x2 9 x 3 Thus, dx x2 9 x2 9 ln sec d = ln sec tan c1 = x x c1 = ln 3 3 ln x = x2 9 x2 9 3 ln 3 c1 = ln x x2 9 c c 1 ln 3 Answer: ln x 2. 3 2x3 2 25 x 2 x2 9 c1 = c dx Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 c , where Since the function f ( x ) 2x3 25 x 2 is continuous on its domain of definition, which is the open interval ( 5 , 5 ) , then it is continuous on the closed interval [ 2 , 3 ] . Thus, the Fundamental Theorem of Calculus can be applied. 2x3 dx using the 25 x 2 technique of Trigonometric Substitution. Then we will apply the Fundamental Theorem of Calculus with one of these antiderivatives. We will first evaluate the indefinite integral Let x 5 sin , where . Then dx 5 cos d . 2 2 2 2 2 2 Also, 25 x 25 25 sin = 25 ( 1 sin ) = 25 cos . 25 x 2 25 cos 2 5 cos 5 cos since cos 0 when . Thus, we have that 2 2 Thus, 2x3 25 x 2 2 ( 125 sin 3 ) ( 5 cos d ) 3 dx = = 250 sin d = 5 cos 250 sin 2 sin d = 250 ( 1 cos 2 ) sin d Let u cos . Then du sin d 250 ( 1 cos 2 ) sin d = 250 ( 1 cos 2 ) ( 1 ) sin d = u3 u3 c = 250 u c = 250 ( 1 u ) du = 250 u 3 3 2 250 250 u ( u 2 3) c = cos ( cos 2 3 ) c 3 3 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 Since x 5 sin , then sin have that cos x . Using right triangle trigonometry, we 5 25 x 2 . 5 5 x 25 x 2 Thus, 250 3 2 3 2x3 25 x 2 25 x 2 5 25 x 2 1 dx = 250 sin 3 d = 250 cos ( cos 2 3 ) c = 3 2 25 x 2 75 50 25 x x 2 50 c = c = 3 1 25 25 25 x 2 50 2 c = ( x 2 50 ) 25 x 2 c 1 3 Now, that we have all the antiderivatives of the integrand f ( x ) 2x3 25 x 2 we can apply the Fundamental Theorem of Calculus to evaluate the definite 3 2x3 dx . integral 2 2 25 x 3 2 2x3 25 x 2 , 3 2 dx = ( x 2 50 ) 25 x 2 = 3 2 2 2 4 ( 59 16 54 21 ) = ( 236 54 21 ) = ( 118 27 21 ) = 3 3 3 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 4 ( 27 21 118 ) 3 Answer: 3. 0 7 4 ( 27 21 118 ) 3 49 t 2 dt Note that we do not have the t in the integral to do the simple substitution to 2 let u 49 t . 2 Since the function f ( t ) 49 t is continuous on its domain of definition, which is the closed interval [ 7 , 7 ] , then it is continuous on the closed interval [ 7 , 0 ] . Thus, the Fundamental Theorem of Calculus can be applied. However, the value of this definite integral can be obtained using the area of 2 the region bounded by the graph of the integrand y 49 t and the horizontal t-axis on the closed interval [ 7 , 0 ] . Since the graph of the 2 function y 49 t is the top-half of the circle given by the equation t 2 y 2 49 . The center of the circle is the origin and the radius is 7. Thus, 2 the region bounded by the graph of y 49 t and the t-axis on the interval [ 7 , 0 ] is the upper left quarter of the circle. Since the area of the 2 circle is given by r and r 7 , then the area of the circle is 49 . Thus, 0 49 49 2 the area of the quarter circle is . Thus, 7 49 t dt = . 4 4 To apply the Fundamental Theorem of Calculus, we will first evaluate the 49 t 2 dt using the technique of Trigonometric Substitution in order to obtain all the antiderivatives of the integrand indefinite integral f (t ) 49 t 2 . Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 Let t 7 sin , where . Then dt 7 cos d . 2 2 2 2 2 2 Also, 49 t 49 49 sin = 49 ( 1 sin ) = 49 cos . 49 t 2 49 cos 2 7 cos 7 cos since cos 0 . Thus, we have that when 2 2 Thus, 49 t 2 dt = ( 7 cos ) ( 7 cos d ) 2 = 49 cos d = sin 2 49 c 4 2 Since t 7 sin , then sin t 1 t sin and . Using right triangle 7 7 trigonometry, we have that cos 49 t 2 . 7 7 t 49 t 2 By the double angle formula for sine, we have that sin 2 2 sin cos . t Thus, sin 2 2 7 49 t 2 2t = 7 49 t 2 49 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 . Thus, sin 2 49 t 2 dt = 49 cos 2 d = 49 c = 4 2 1 1 2 t 1 t 49 sin 2 7 4 1 1 t 1 t 49 sin 2 7 2 sin 1 t t 7 49 2 49 t 2 c = 49 49 t 2 c = 49 49 t 2 c 49 2 Now, that we have all the antiderivatives of the integrand f ( t ) 49 t , we can apply the Fundamental Theorem of Calculus to evaluate the definite integral 0 7 49 t 2 dt . 0 0 7 49 t 2 t 49 t 2 49 1 t sin dt = = 2 7 49 7 49 49 [ sin 1 0 0 ( sin 1 ( 1) 0 ) ] = [ sin 1 0 sin 1 ( 1) ] = 2 2 49 2 0 2 49 49 = = 2 2 4 49 Answer: 4 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 4. 6x3 8 x 16 2 5 dx Since the function f ( x ) 6x3 is continuous on its domain of x 2 16 definition, which is the set ( , 4 ) ( 4 , ) , then it is continuous on the closed interval [ 5 , 8 ] . Thus, the Fundamental Theorem of Calculus can be applied. We will first evaluate the indefinite integral 6x3 dx using the x 16 technique of Trigonometric Substitution. Then we will apply the Fundamental Theorem of Calculus with one of these antiderivatives. Let x 4 sec , where 0 dx 4 sec tan d . 2 3 or . Then 2 2 2 2 2 2 Also, x 16 16 sec 16 = 16 ( sec 1) = 16 tan . x 2 16 16 tan 2 4 tan 4 tan since tan 0 3 when 0 or . Thus, we have that 2 2 Thus, 6 ( 64 sec 3 ) ( 4 sec tan d ) 4 dx = 384 sec d = = 4 tan x 2 16 6x3 384 sec 2 sec 2 d = 384 ( 1 tan 2 ) sec 2 d 2 Let u tan . Then du sec d u3 c = 384 ( 1 tan ) sec d = 384 ( 1 u ) du = 384 u 3 2 2 2 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 384 u ( u 2 3 ) c = 128 tan ( tan 2 3 ) c 3 Since x 4 sec , then sec have that tan x . Using right triangle trigonometry, we 4 x 2 16 . 4 x x 2 16 4 6x3 Thus, 128 x 2 16 x 2 16 48 c = 32 4 16 16 2 x 16 2 dx = 384 sec 4 d = 128 tan ( tan 2 3 ) c = x 2 16 x 2 32 c = 1 16 x 2 16 x 2 32 c = 2 ( x 2 32 ) 1 1 x 2 16 c Now, that we have all the antiderivatives of the integrand f ( x ) 6x3 x 2 16 we can apply the Fundamental Theorem of Calculus to evaluate the definite 8 6x3 dx . integral 5 2 x 16 8 5 6x3 x 2 16 dx = 2 ( x 2 32 ) x 16 2 8 5 = Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 , 2 ( 96 48 57 9 ) = 2 ( 96 16 3 171 ) = 2 ( 384 3 171 ) = 768 3 342 Answer: 768 3 342 5. y 5 ( 9 7 y 2 ) 3 / 2 dy We want to rewrite the integral before we use the technique of Trigonometric Substitution. y y 5 5 ( 9 7 y 2 ) 3 / 2 dy = 7 3/ 2 Let y 9 2 y 7 3 7 9 y 7 y2 7 3/ 2 5 3/ 2 dy = 7 3/ 2 dy = 9 y y2 7 3/ 2 5 dy . Then dy 2 2 tan , where 3 7 sec 2 d . 9 9 9 2 9 9 2 2 2 y tan ( 1 tan ) Also, = = sec . 7 7 7 7 7 9 2 Thus, y 7 3/ 2 9 sec 2 7 since sec 1 0 when y 5 ( 9 7 y 2 ) 3 / 2 dy = 7 3/ 2 27 sec 7 3/ 2 3 27 sec 3 3/ 2 7 . Thus, we have that 2 2 3/ 2 9 y y2 7 3/ 2 5 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 dy = 7 3/ 2 35 tan 5 5/ 2 7 7 3/ 2 3 9 7 9/2 39 = 3 7 39 73 ( sec 3 1 / 2 sec 2 d = 7 39 tan sec d = 3 7 5 tan 39 73 33 3 / 2 sec 3 7 4 5 tan 5 sec 5 d = sec 4 sec tan d = ( tan 2 ) 2 sec 4 sec tan d = 2 1 ) 2 sec 4 sec tan d Let u sec . Then du sec tan d 39 73 39 ( sec 1 ) sec sec tan d = 3 7 39 73 (u 39 73 u9 2u 7 u5 c = 9 7 5 2 4 2 2u 2 4 39 1 ) u du = 3 7 4 (u 8 (u 2 1 ) 2 u 4 du = 2 u 6 u 4 ) du = 39 1 2 1 sec 9 sec 7 sec 5 c 3 7 9 7 5 Since y 3 7 tan , then tan trigonometry, we have that sec 7 y . Using right triangle 3 9 7y2 3 . Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 9 7y2 7y 3 Thus, y 5 ( 9 7 y 2 ) 3 / 2 dy = 7 9 y y2 7 5 3/ 2 dy = 39 1 2 1 tan sec d = 3 sec 9 sec 7 sec 5 c = 7 9 7 5 39 73 39 73 1 ( 9 7 y 2 ) 9/2 2 ( 9 7 y 2 ) 7/2 1 ( 9 7 y 2 ) 5/ 2 9 7 9 3 7 3 5 35 c = 39 1 1 ( 9 7 y 2 ) 9/2 2 ( 9 7 y 2 ) 7/2 1 ( 9 7 y 2 ) 5/ 2 7 3 3 5 9 34 7 32 5 1 c = 34 73 5 5 1 ( 9 7 y 2 ) 9/2 2 ( 9 7 y 2 ) 7/2 ( 9 7 y 2 ) 5/ 2 9 81 7 9 5 81 343 ( 9 7 y 2 ) 9/2 2( 9 7 y 2 ) 7/2 ( 9 7 y 2 ) 5/ 2 729 63 5 81 Answer: 343 6. 3/ 2 5x c = c ( 9 7 y 2 ) 9/2 2( 9 7 y 2 ) 7/2 ( 9 7 y 2 ) 5/ 2 729 63 5 4 x 2 11 dx Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 c This integral can be evaluated by simple substitution which is FASTER than trigonometric substitution. 2 Let u 4 x 11 Then du 8 x dx 5x 5 8 4 x 2 11 dx = 5 x u du = 5 8 u 1 / 2 du = 4 x 2 11 dx = 5 8 8x 4 x 2 11 dx = 5 2 3/ 2 5 u c = ( 4 x 2 11 ) 3 / 2 c 8 3 12 If you missed the simple substitution and proceeded with the technique of Trigonometric Substitution, then your work would have been the following (and the following, and the following.) 5x 4 x 2 11 dx = 11 Let x dx 2 11 2 2 Also, x 5x 11 4 x 2 dx = 10 x 4 sec , where 0 x2 11 dx 4 3 or . Then 2 2 sec tan d . 11 11 11 11 11 sec 2 ( sec 2 1 ) = tan 2 . = 4 4 4 4 4 11 11 11 2 tan tan tan since Thus, 4 2 2 3 tan 0 when 0 or . Thus, we have that 2 2 x2 5x 11 4 4 x 2 11 dx = 10 x x2 11 dx = 4 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 10 11 2 55 11 4 sec tan 2 11 2 tan 11 2 sec tan d = sec 2 d 2 Let u tan . Then du sec d 55 11 4 55 11 12 tan 2 sec d = 2 55 11 4 u 2 du = 55 11 tan 3 c Since x 11 2 sec , then sec trigonometry, we have that tan 2x 11 . Using right triangle 4 x 2 11 11 . 2x 4 x 2 11 11 5x 55 11 Thus, 4 4 u3 c = 3 4 x 2 11 dx = 10 x tan 2 sec 2 d = 55 11 12 x2 11 dx = 4 tan 3 c = Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 55 11 12 ( 4 x 2 11 ) 3 / 2 11 11 c = 5 ( 4 x 2 11 ) 3 / 2 c 12 5 ( 4 x 2 11 ) 3 / 2 c Answer: 12 7. t 3 ( 6 t 2 ) 5 / 2 dt 6 sin , where Let t . Then dt 2 2 6 cos d . 2 2 2 2 Also, 6 t 6 6 sin = 6 ( 1 sin ) = 6 cos . 2 5/ 2 ( 6 cos 2 ) 5 / 2 6 5 / 2 cos Thus, ( 6 t ) cos 0 when t 3 sin 6 9/2 3 6 5 / 2 cos 5 since . Thus, we have that 2 2 ( 6 t 2 ) 5 / 2 dt = 6 9/2 5 6 3/ 2 sin 3 ( 6 5 / 2 cos 5 ) 6 1 / 2 cos d = cos 6 d = 6 9 / 2 sin 2 cos 6 sin d = ( 1 cos 2 ) cos 6 sin d Let u cos . Then du sin d 6 9/2 (1 cos 2 ) cos 6 sin d = 6 9/2 (1 cos 2 ) cos 6 ( 1) sin d = 6 9 / 2 6 9/2 (u u 8 ) du = 6 9 / 2 6 (u 8 u 6 ) du = Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 (1 u 2 ) u 6 du = 6 9/2 u9 u7 1 1 c = 6 9 / 2 cos 9 cos 7 c 7 7 9 9 Since t t 6 sin , then sin we have that cos . Using right triangle trigonometry, 6 6 t2 . 6 6 t 6 t2 Thus, t 3 ( 6 t 2 ) 5 / 2 dt = 6 9 / 2 sin 3 cos 6 d = 1 1 6 9 / 2 cos 9 cos 7 c = 7 9 6 9/2 1 ( 6 t 2 ) 9/2 1 ( 6 t 2 ) 7/2 9/2 9 6 7 6 7/2 c = 1 6 ( 6 t 2 ) 9/2 ( 6 t 2 ) 7/2 c 9 7 Answer: 1 6 ( 6 t 2 ) 9/2 ( 6 t 2 ) 7/2 c 9 7 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 8. (9 y 2 dy 25 ) 3 / 2 Note that we do not have the y in the integral to do the simple substitution to 2 let u 9 y 25 . We will use the technique of Trigonometric Substitution. First, we will need to rewrite the integral. (9 y 1 27 2 dy 25 2 9 y 9 3/ 2 = dy 9 3/ 2 25 2 y 9 3/ 2 dy y Let y dy dy = 25 ) 3 / 2 25 9 2 3/ 2 3 5 sec , where 0 or . Then 3 2 2 5 sec tan d . 3 2 Also, y 25 25 25 25 25 sec 2 ( sec 2 1 ) = tan 2 . = 9 9 9 9 9 25 2 Thus, y 9 3/ 2 3/ 2 25 3 / 2 125 3 25 2 tan 3 / 2 tan tan 3 9 27 9 3 since tan 0 when 0 or . Thus, we have that 2 2 (9 y 2 1 dy = 27 25 ) 3 / 2 y dy 2 25 9 3/ 2 = Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 = 1 27 5 sec tan d 1 5 1 3 = 27 3 125 125 3 tan 27 27 1 5 27 27 3 125 1 75 sec d = tan 2 1 1 1 sec cot d = 1 3 25 2 1 cos 2 d = cos sin 2 cos d sin 2 Let u sin . Then du cos d 1 75 cos 1 d = 2 sin 75 du 1 2 = u 75 u 2 1 u 1 c = du = 75 1 1 1 1 1 1 c = c = csc c 75 sin 75 u 75 5 3y sec , then sec . Using right triangle trigonometry, 3 5 3y csc we have that . 9 y 2 25 Since y 3y 9 y 2 25 5 Thus, (9 y 2 dy 1 = 25 ) 3 / 2 75 cos 1 d csc c = = sin 2 75 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 1 75 3y 9y Answer: 9. 2 c = 25 y 25 9y 25 2 ( x 2 4) 2 dx 3x ( x 2 4) 2 1 dx = 3x 3 1 3 y 25 9y 2 25 c c x 4 8 x 2 16 dx = x 1 x4 16 3 4 x 2 16 ln x c = x 8x dx = 3 4 x 1 4 4 16 x x2 ln x c 12 3 3 If you evaluate the integral using the technique of Trigonometric Substitution, then your work would have been the following. Let x 2 tan , where . Then dx 2 sec 2 d . 2 2 2 2 2 2 Also, x 4 4 tan 4 = 4 ( tan 1 ) = 4 sec . 2 2 2 2 4 Thus, ( x 4 ) ( 4 sec ) 16 sec . Thus, we have that ( x 2 4) 2 dx = 3x 16 sec 4 ( 2 sec 2 d ) 16 = 3 ( 2 tan ) 3 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 ( tan ) 1 sec 6 d = 16 3 16 3 ( tan ) 16 3 ( tan ) 1 sec 4 sec 2 d = 1 ( tan ) ( sec 2 ) 2 sec 2 d 1 ( 1 tan 2 ) 2 sec 2 d 2 Let u tan . Then du sec d 16 3 ( tan ) 1 ( 1 tan 2 ) 2 sec 2 d = 16 3 u 1 ( 1 2 u 2 u 4 ) du = 16 3 u4 u 2 ln u c = 4 16 3 (u 3 16 3 u 1 ( 1 u 2 ) 2 du = 2 u u 1 ) du = 16 1 4 2 tan tan ln tan c 3 4 Since x 2 tan , then tan Thus, ( x 2 4) 2 16 dx = 3x 3 x . 2 ( tan ) 1 sec 6 d = 16 1 4 2 tan tan ln tan c 1 = 3 4 16 3 1 x4 x2 x ln 4 2 4 16 c 1 = 16 3 x4 x2 ln x ln 2 c 1 = 4 64 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 1 4 4 2 16 16 x x ln x ln 2 c 1 = 12 3 3 3 1 4 4 2 16 16 x x ln x c , where c c 1 ln 2 12 3 3 3 1 4 4 16 x x2 ln x c 12 3 3 Answer: Example If a is a positive real number, b is any real number, m is a rational number, and n is an odd integer, then determine the trigonometric integral for the integral bx m ( a 2 x 2 ) n / 2 dx if you do the trigonometric substitution to let x a sin , where . 2 2 Since x a sin , then dx a cos d . a 2 x 2 = a 2 ( a sin ) 2 = a 2 a 2 sin 2 = a 2 (1 sin 2 ) = a 2 cos 2 . 2 2 n n 2 2 n/2 2 2 n/2 Thus, ( a x ) = ( a cos ) = ( a cos ) = ( a cos ) = n n a cos . Since a 0 , then a a . Since cos 0 whenever , 2 2 2 2 n/2 n n then cos cos . Thus, ( a x ) = a cos . Thus, bx (a x b sin cos m am n 1 Answer: 2 m bx m 2 ) n / 2 dx = b a m sin m ( a n cos n ) a cos d = n 1 d . ( a 2 x 2 ) n / 2 dx = a m n 1 b sin m cos n 1 d For example, in our Example 2 above, we had the integral Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 3 2x3 2 25 x 2 dx . For this integral, a 5 , b 2 , m 3 , and n 1 . Thus, 5 3 2 sin 3 d = 250 sin 3 2x3 25 x 2 dx = d . Example If a is a positive real number, b is any real number, m is a rational number, and n is an integer, then determine the trigonometric integral for the integral bx m ( a 2 x 2 ) n / 2 dx if you do the trigonometric substitution to let x a tan , where . 2 2 2 Since x a tan , then dx a sec d . a 2 x 2 = a 2 ( a tan ) 2 = a 2 a 2 tan 2 = a 2 (1 tan 2 ) = a 2 sec 2 . 2 2 n n 2 2 n/2 2 2 n/2 Thus, ( a x ) = ( a sec ) = ( a sec ) = ( a sec ) = sec . Since a 0 , then a a . Since sec 1 0 whenever , then sec sec . Thus, ( a 2 x 2 ) n / 2 = a n sec n . 2 2 a n Thus, n b x (a x b tan sec m a m n 1 Answer: 2 m bx m 2 ) n / 2 dx = b a m tan m ( a n sec n ) a sec 2 d = n2 d . ( a 2 x 2 ) n / 2 dx = a m n 1 b tan m sec n 2 d For example, in our Example 1 above, we had the integral integral, a 3 , b 1 , m 0 , and n 1 . Thus, 3 0 (1) tan 0 sec d = sec d . Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 dx x2 9 dx = x2 9 . For this Example If a is a positive real number, b is any real number, m is a rational number, and n is an odd integer, then determine the trigonometric integral for the integral bx m ( x 2 a 2 ) n / 2 dx if you do the trigonometric substitution to let x a sin , where 0 3 or . 2 2 Since x a sec , then dx a sec tan d . x 2 a 2 = ( a sec ) 2 a 2 = a 2 sec 2 a 2 = a 2 ( sec 2 1) = a 2 tan 2 . 2 2 n n 2 2 n/2 2 2 n/2 Thus, ( x a ) = ( a tan ) = ( a tan ) = ( a tan ) = a n n tan . Since a 0 , then a a . Since tan 0 whenever 0 or Thus, 3 2 2 n/2 n n , then tan tan . Thus, ( x a ) = a tan . 2 2 b x m ( x 2 a 2 ) n / 2 dx = b a m sec m ( a n tan n ) a sec tan d = a m n 1 b tan n 1 sec m 1 d . Answer: bx m ( x 2 a 2 ) n / 2 dx = a m n 1 b tan n 1 sec m 1 d For example, in our Example 8 above, we had the integral dy 25 2 9 y 9 the integral 3/ 2 = dy 9 3/ 2 dy 25 27 y 2 9 3/ 2 25 2 y 9 ,a 3/ 2 = (9 y 2 dy = 25 ) 3 / 2 dy 25 27 y 2 9 3/ 2 . For 5 1 ,b , m 0 , and n 3 . Thus, 3 27 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 dy 3/ 2 5 = 3 2 25 27 y 2 9 1 9 1 2 cot sec d = 75 25 27 1 27 ( tan ) 2 sec d = 1 cos 2 1 d = 75 sin 2 cos Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1860 cos d sin 2