<p> Calculus BC Sections 5/6 – 5/8 Review Worksheet answers</p><p>Find the derivative. 1 = sinh x cosh x 1. y = arcsin(4x) sinh x dy 1 d(4x) = =  cosh x dx 1 (4x) 2 dx = tanh x 4 = 6. y = sinh x ∙ ecosh x 116x 2 dy cosh x  3  = sinh x ∙ (e ∙sinh x) 2. y = arctan  dx  x  + ecosh x ∙cosh x 1 d(3x 1 ) = sinh2x ∙ ecosh x + cosh x ∙ ecosh x  cosh x 2 dy 2 = e (sinh x + cosh x) =  3  dx dx 1    x  7. y = 2x∙arctan x – ln(x2 + 1) 1  3  dy  1  1 2 = 2x   arctan x  2   2x = 9 x dx  x 2 1  x 2 1 1 2 x 2x 2x  3 = 2  2  arctan x  2 x 1 x 1 = 2 x  9 = 2∙arctan x</p><p>3. y = arcsec(2x+1) Find the indefinite integral. 1 d(2x 1) dy  = 2 dx 1 dx 2x 1  2x 1 1  dx 8. I =  2 1 4  9x =  2 1 2x 1  4x 2  4x  dx =  2 2 2 (2)  (3x) = 3x = 2 2x 1  2 x 2  x 3∙dx = 2d 1 dx = 2/3d = 2 1 2 2x 1  x  x  dx I =  2 2 (2)  (2 ) 3 2 2 4. y = sinh x + cosh x 2 1  dx dy =  2 = 2∙sinhx∙coshx + 2∙coshx∙sinhx 3 4  4 dx 2 1 1 = 4∙sinh x∙cosh x   dx =  2 = 2(2∙sinh x∙cosh x) 3 2 1 = 2∙sinh(2x) 1 = ∙arcsin() + C 3 5. y = ln(cosh x) Since 3x = 2,  = 3x/2 dy 1 d(cosh x) =  1  3x  dx cosh x dx I = = ∙arcsin  + C 3  2  Calculus BC Sections 5/6 – 5/8 Review Worksheet answers</p><p>1 1 9. I =  dx 11. I =  dx  25x 2  4  x 2  4x 13 1 Completing the square: =  dx 2 2  (5x)2  22 x + 4x + 13 = (x + 4x ) + 13 = (x2 + 4x + 22 ) + 13 – 22 5x = 2 = (x + 2)2 + 9 5dx = 2d = (x + 2)2 + (3)2 dx = 2/5 d 1 1 2 I =  dx I =  dx  (x  2) 2  (3) 2  (2 )2  22 5 x + 2 = 3 2 1 =  dx dx = 3d  2 5 4  4 1 2 1 1 I = 3d =   dx  (3 )2  (3)2 5 4   2 1 1 1 = 3  d = ∙arctan() + C  9 2  9 10 1 1 Since 5x = 2,  = 5x/2. = 3  d 9   2 1 1  5x  I = ∙arctan  + C 1 10  2  = ∙arctan() + C 3 Since x + 2 = 3,  = (x+2)/3 1  dx 1  x  2  10. I =  2 4x 16x  9 I = = ∙arctan  + C 3  3  1  dx =  2 4x 4x  (3) 2 1    dx 12. I =  2 4x = 3 40  6x  x 4dx = 3d Completing the square: dx = 3/4 d 40 – 6x – x2 = -(x2 + 6x – 40) 1 3 = -[(x2 + 6x ) – 40]  dx 2 2 2 I =  2 = -[(x + 6x + 3 ) – 40 - 3 ] 3 3  (3)2 4   = -[ (x+3)2 – 49] 3 1 1 2 2   dx = -[ (x+3) – (7) ] =  2 2 2 4 3  9  9 = (7) – (x+3) 1 3 1 1 1  dx    dx I =  2 2 =  2 4 3 3   1 (7)  (x  3) 1 x + 3 = 7 = ∙arcsec|| + C 12 dx = 7d 1 4x  7d Since 4x = 3,  = I =  2 2 3 (7)  (7 ) 1 4x 1 I = ∙arcsec + C 7  dx =  2 12 3 49  49 Calculus BC Sections 5/6 – 5/8 Review Worksheet answers</p><p>1 1 1 1 7   dx  2d =  2 I = 0 2 2 7 1 (2 )  (2) = 1∙arcsin() + C 1 1 = 2  d Since x + 3 = 7,  = (x + 3)/7 0 4 2  4  x  3  1 1 1 I = arcsin  + C = 2   d  7  4 0  2 1 1 = arctan 1 sinh x 2 0 13. I =  dx  cosh x 1 = ½ [arctan(1) – arctan(0)] 1 1   1    = sinh x  dx =  0    cosh x 1 2  4  2  4  8 u = coshx + 1 du = sinhx∙dx 0.5 12  dx 1 16. I = 0.2 2 I =  du 1 4x  u 0.5 1 = ln | u | + C 12  dx = 0.2 2 2 = ln(cosh x + 1) + C (1)  (2x) 2x =  When x = 0.5,  = 1 sinh1  2dx = d When x = 0.2,  = 0.4 14. I = x  dx  x 2 dx = ½ d 1 2 1 1  1  =  sinh(x )  x  dx 12  dx = 0.4 2 u = x-1 1  2  du = -1∙x-2∙dx 1 1 -2 6  dx = 0.4 2 -du = x ∙dx 1 I = sinh(u)(du) 1  = 6arcsin 0.4 =   sinh(u)  du = 6[ arcsin(1) – arcsin(0.4) ] = -cosh(u) + C = 6.9557 1   ln(2) = -cosh  + C 17. I = [sinh(x)  cosh(x)]dx  x  0 ln(2) =cosh x  sinh x0 Evaluate the definite integral. ln(2) e x  e x e x  e x  =  2 1    2 2  0 15. I = 2  dx 0 x  4 ln(2) 2e x  2 1 =   =  dx 2 0 (x)2  (2) 2   0 ln(2) x = 2 When x = 2,  = 1 = e x dx = 2d When x = 0,  = 0  0 = eln(2) – e0 = 2 – 1 = 1</p>