Make trigonometric substitutions to evaluate the

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18. Make trigonometric substitutions to evaluate the following integrals: (a)(b)(c)Solution:(a) Substitutex=8 sinθ,dx=8 cosθ dθ. Whenx= 0,θ= 0,and whenx= 2,θ= arcsin(2/8) = arcsin(2/22) = arcsin(1/2) = Z 2 0 16 (8 - x 2 ) 3 / 2 dx Z 1 0 18 (9 + 16 x 2 ) 3 / 2 dx Z 4 t 2 t 2 - 4 dt
π/ 4. Hence the integral becomes Z π/ 4 0 16 (8 - 8 sin 2 θ ) 3 / 2 8 cos θ dθ = Z π/ 4 0 16 8 8 8 cos 3 θ cos θ dθ = Z π/ 4 0 16 8 8 8 cos 3 θ cos θ dθ = 2 Z π/ 4 0 sec 2 θ dθ = 2 tan θ π/ 4 0 = 2 . (b) Substitute x = (3 / 4) tan θ , dx = (3 / 4) sec 2 θ dθ . When x = 0, θ = 0, and when x = 1, tan θ = 4 / 3 so θ = arctan(4 / 3). The integral therefore becomes Z arctan(4 / 3) 0 18 (9 + 9 tan 2 θ ) 3 / 2 3 4 sec 2 θ dθ = 1 2 Z arctan(4 / 3) 0 sec 2 θ sec 3 θ = 1 2 Z arctan(4 / 3) 0 cos θ dθ = 1 2 sin θ arctan(4 / 3) 0 By looking at a right-angle triangle with sides 3 , 4 , 5, we see that sin(arctan(4 / 3)) = 4 / 5, so the value of the integral is 2 / 5. (c) Substitute t = 2 sec θ , dt = 2 sec θ tan θ dθ . Then t 2 t 2 - 4 = 4 sec 2 θ 4 sec 2 θ - 4 = 8 sec 2 θ tan θ , so the integral becomes R 8 sec θ tan θ 8 sec 2 θ tan θ = R 1 sec θ = R cos θ dθ = sin θ + C . Then sin θ = 1 - cos 2 θ = p 1 - (1 / sec θ ) 2 = p 1 - 4 /t 2 (as sin θ is positive on the range of sec - 1 ), so the integral is p 1 - 4 /t 2 + C . 19. Use an appropriate substitution to find Z 2 0 4 4 + x 2 3 / 2 dx Z 1 (4 x 2 - 9) 3 / 2 dx for x > 3 / 2. Hint: try x = (3 / 2) sec θ ; you will eventually need to write sin θ as a function of x . Solution: (a) Let x = 2 tan θ ; dx = 2 sec 2 θ dθ and 4 + x 2 = 4(1 + tan 2 θ ) = 4 sec 2 θ . When x = 0 , θ = 0 and x = 2 = 2 tan θ = 2 = tan θ = 1 = θ = π 4 .
R 2 0 θ ] π/ 4 0 = 2 . (b) Let x = (3 / 2) sec θ ; dx = (3 / 2) sec θ tan θ dθ and 4 x 2 - 9 = 9(sec 2 θ - 1) = 9 tan 2 θ . Thus (4 x 2 - 9) 3 / 2 = 27 tan 3 θ . R 1 (4 x 2 - 9) 3 / 2 dx = R (3 / 2) sec θ tan θ 27 tan 3 θ = 1 18 R sec θ tan 2 θ = 1 18 R cos θ sin 2 θ = - 1 18 1 sin θ . The last integral was evaluated by using the substitu- tion, u = sin θ ; then du = cos θ dθ and R cos θ sin 2 θ = R 1 u 2 du = - 1 u + c . Now x = (3 / 2) sec θ = x = 3 2 cos θ = cos θ = 3 2 x = sin θ = q 1 - ( 3 2 x ) 2 = q 4 x 2 - 9 4 x 2 = 1 2 x 4 x 2 - 9. Thus R 1 (4 x 2 - 9) 3 / 2 dx = - 1 18 2 x 4 x 2 - 9 + c = - x 9 1 4 x 2 - 9 + c . 20. Use the given substitution to evaluate the integral: Z sec θ dθ ; Z dx x 2 + a 2 . a 2 + x 2 , so the answer is ln | 1 a a 2 + x 2 + x a | + C . 21. Evaluate the following integrals by completing the square and makingsubstitutions: Z 2 1 2 x x 2 - 4 x + 5 dx Z p 4 x 2 + 4 x - 3 dx Z ( x - x 2 ) - 3 / 2 dx .
Thus the integral becomes R 0 - π/ 4 2(tan t + 2) dt = (2 ln | sec t | + 4 t ) 0 - π/ 4 = 0 - (ln 2 - π ) = π - ln 2. (b) 4 x 2 + 4 x - 3 = 4( x + 1 / 2) 2 - 4 so subs x + 1 / 2 = cosh t, dx = sinh t dt and integral becomes R 2 sinh 2 t dt = R ( - 1+cosh 2 t ) dt = sinh 2 t/ 2 - t + C = sinh t cosh t - t + C = p ( x + 1 / 2) 2 - 1 ( x + 1 / 2) - cosh - 1 ( x + 1 / 2) + C . (c) x - x 2 = - ( x - 1 / 2) 2 + 1 / 4 so subs x - 1 / 2 = (1 / 2) sin t , dx = (1 / 2) cos t dt and the integral becomes R [(1 / 2) cos t ] - 3 (1 / 2) cos t dt = R 4 sec 2 t dt = 4 tan t + C . 2 x - 1 = sin t cos t = p 1 - (2 x - 1) 2 = 2 x - x 2 , tan t = (2 x - 1) / 2 x - x 2 . Hence the integral is 4 x - 2 x - x 2 + C .

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