University:
University of Central Florida
Course:
MAC 2312 | Calculus with Analytic Geometry II
Academic year:

2023
Views:

21

Pages:

3
Author:

Maria

Q1: S(tan (3x)) (-2sec (3x)) dx = S(rtam³ (3x)) (-27㎡(sec² (3x) sec²(3x))) dx = S(tan (3x)) (-27(tan²(3x) +1) sec²(3x)) dx u=tan (3x) du=3sec²(5x) 3 f(x(u²)) (-2m (u²+1))du= -20m² f(u² (u²+1)) du 27 3 -2² justu du: -2² (+u+u")+c 3 C 3 = -20²[(tan (3x))+(tan" (3x))] +C 6 Q2: S (tom² (x) (-see'(x)) dx 0 pf ((tam²(x))² sec² (x) sec(x)tan(x)) dx 0 (((sec² (x)-1)² sec²(x)sec(x)tan(x)) dx u = sec(x) du sec(x) tam (x) dx √2 √2 = 52 (u² - 1)² u² duz ve uy - zu² + 1) u du S 2 4 2 2u+ pvz u6 - 244 + u² du = 1 + Lu 3 √2 1 7 S =(=(√2)²-==(√2)²++=(√2)²)-((1)²-(1)²++ (1)²) S 7 S 3 3 3 3 Q3. S-3 cos² (2x) sin² (2x)dx COS S 3 = S(-3m (1-sin² (2x)) cos (2x)) (sin² (2x))dx u= Sin (2x) du: 2 COS (2x) dx „S-3m (1-4²) u² dum = (-3² + B²) du 2π +C = -34%+ 46 46 + C 4 8 6 2 2 4 = -3 sin" (200x) + sin² (2x) + C 6 8 4 + 3 Q4: S-sin² (2x)dx = S-ㅠ (1-cos²(2x)) sin (2x) dx u=cos (2x) du-2 sin (2x)dx =S-TCI-COS 3-ㅠ (1-cos² (2x)) sin (2x)dx=-1+f-ㅠ (1-u²) du 2 3 =(u) + c = -² + C = cos(2x) - Tcos² (2x) +e 3 3 πα πα 2 te 2 6 2 6 3 Q5: S-27 tans (x) sec² (ix) dx =S-2 (tam² (x))² sec²(x) sec (Tx) tan (Tx) dx こ =S-2 (Sec² (x)-1)² sec² (x) sec (Tx)tan(x)dx 2 4= sec(x) du: Sec (Tx) fan (x) dx S-2 ((u²-1)² u²) dhe = -2 (u4-2u² +1) u² du --2ju-Zu²+u² du=-2[+ut-u²+ ½²]+c 6- =-2[sec(x) S tu 3 2 sec (πx) 7 3 + 3