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# Waleed

• Date Submitted: 03/31/2015 06:31 AM
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Unit 5. Integration techniques
5A. Inverse trigonometric functions; Hyperbolic functions
âˆš Ï€ 3 b) sin ( )= 5A-1 a) tan 2 3 âˆš âˆš âˆš c) tan Î¸ = 5 implies sin Î¸ = 5/ 26, cos Î¸ = 1/ 26, cot Î¸ = 1/5, csc Î¸ = 26/5, âˆš sec Î¸ = 26 (from triangle) âˆš Ï€ Ï€ Ï€ 3 Ï€ )= e) tanâˆ’1 tan( ) = d) sinâˆ’1 cos( ) = sinâˆ’1 ( 6 2 3 3 3 âˆ’Ï€ âˆ’Ï€ âˆ’Ï€ 2Ï€ )= g) lim tanâˆ’1 x = . f) tanâˆ’1 tan( ) = tanâˆ’1 tan( xâ†’âˆ’âˆž 3 3 3 2
âˆ’1

âˆš Ï€ 3= 3

âˆ’1

2

5A-2 a)
1 2b

dx = tanâˆ’1 x x2 + 1 dx = x2 + b2
2b b

2 1

= tanâˆ’1 2 âˆ’

Ï€ 4
2 1

b)
b 1

d(by) (put x = by) = (by)2 + b2
1 âˆ’1

1 Ï€ dy = (tanâˆ’1 2 âˆ’ ) b(y 2 + 1) b 4

c)
âˆ’1

dx âˆš = sinâˆ’1 x 1 âˆ’ x2

=

Ï€ âˆ’Ï€ âˆ’ =Ï€ 2 2 1 1âˆ’ y2 = (x + 1) âˆš . Hence 2 x

5A-3 a) y =

xâˆ’1 , so 1 âˆ’ y 2 = 4x/(x + 1)2 , and x+1

dy 2 = dx (x + 1)2 d dy/dx sinâˆ’1 y = dx 1 âˆ’ y2 2 (x + 1) = Â· âˆš (x + 1)2 2 x 1 âˆš = (x + 1) x b) sech2 x = 1/ cosh2 x = 4/(ex + eâˆ’x )2 âˆš âˆš c) y = x + x2 + 1, dy/dx = 1 + x/ x2 + 1. âˆš d 1 dy/dx 1 + x/ x2 + 1 âˆš = âˆš ln y = = dx y x + x2 + 1 x2 + 1 d) cos y = x =â‡’ (âˆ’ sin y)(dy/dx) = 1 dy âˆ’1 âˆ’1 = =âˆš dx sin y 1 âˆ’ x2 e) Chain rule: d sinâˆ’1 (x/a) = dx 1 1 âˆ’ (x/a)2 Â· 1 1 = âˆš 2 âˆ’ x2 a a

S. SOLUTIONS TO 18.01 EXERCISES

f) Chain rule: d sinâˆ’1 (a/x) = dx 1 1 âˆ’ (a/x)2 Â· âˆ’a âˆ’a = âˆš x2 x x2 âˆ’ a2

âˆš g) y = x/ 1 âˆ’ x2 , dy/dx = (1 âˆ’ x2 )âˆ’3/2 , 1 + y 2 = 1/(1 âˆ’ x2 ). Thus d 1 dy/dx = (1 âˆ’ x2 )âˆ’3/2 (1 âˆ’ x2 ) = âˆš tanâˆ’1 y = dx 1 + y2 1 âˆ’ x2 Why is this the same as the derivative of sinâˆ’1 x? âˆš âˆš h) y = 1 âˆ’ x, dy/dx = âˆ’1/2 1 âˆ’ x, 1 âˆ’ y 2 = x. Thus, d dy/dx = sinâˆ’1 y = dx 2 1 âˆ’ y2 x(1 âˆ’ x) âˆ’1

5A-4 a) y â€² = sinh x. A tangent line through the origin has the equation y = mx. If it meets the graph at x = a, then ma = cosh(a) and m = sinh(a). Therefore, a sinh(a) = cosh(a) . b) Take the diï¬€erence: F (a) = a sinh(a) âˆ’ cosh(a) Newtonâ€™s method for ï¬nding F (a) = 0, is the iteration an+1 = an âˆ’ F (an )/F â€²...