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  • Date Submitted: 03/31/2015 06:31 AM
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Unit 5. Integration techniques
5A. Inverse trigonometric functions; Hyperbolic functions
√ π 3 b) sin ( )= 5A-1 a) tan 2 3 √ √ √ c) tan θ = 5 implies sin θ = 5/ 26, cos θ = 1/ 26, cot θ = 1/5, csc θ = 26/5, √ sec θ = 26 (from triangle) √ π π π 3 π )= e) tan−1 tan( ) = d) sin−1 cos( ) = sin−1 ( 6 2 3 3 3 −π −π −π 2π )= g) lim tan−1 x = . f) tan−1 tan( ) = tan−1 tan( x→−∞ 3 3 3 2
−1

√ π 3= 3

−1

2

5A-2 a)
1 2b

dx = tan−1 x x2 + 1 dx = x2 + b2
2b b

2 1

= tan−1 2 −

Ï€ 4
2 1

b)
b 1

d(by) (put x = by) = (by)2 + b2
1 −1

1 π dy = (tan−1 2 − ) b(y 2 + 1) b 4

c)
−1

dx √ = sin−1 x 1 − x2

=

π −π − =π 2 2 1 1− y2 = (x + 1) √ . Hence 2 x

5A-3 a) y =

x−1 , so 1 − y 2 = 4x/(x + 1)2 , and x+1

dy 2 = dx (x + 1)2 d dy/dx sin−1 y = dx 1 − y2 2 (x + 1) = · √ (x + 1)2 2 x 1 √ = (x + 1) x b) sech2 x = 1/ cosh2 x = 4/(ex + e−x )2 √ √ c) y = x + x2 + 1, dy/dx = 1 + x/ x2 + 1. √ d 1 dy/dx 1 + x/ x2 + 1 √ = √ ln y = = dx y x + x2 + 1 x2 + 1 d) cos y = x =⇒ (− sin y)(dy/dx) = 1 dy −1 −1 = =√ dx sin y 1 − x2 e) Chain rule: d sin−1 (x/a) = dx 1 1 − (x/a)2 · 1 1 = √ 2 − x2 a a

S. SOLUTIONS TO 18.01 EXERCISES

f) Chain rule: d sin−1 (a/x) = dx 1 1 − (a/x)2 · −a −a = √ x2 x x2 − a2

√ g) y = x/ 1 − x2 , dy/dx = (1 − x2 )−3/2 , 1 + y 2 = 1/(1 − x2 ). Thus d 1 dy/dx = (1 − x2 )−3/2 (1 − x2 ) = √ tan−1 y = dx 1 + y2 1 − x2 Why is this the same as the derivative of sin−1 x? √ √ h) y = 1 − x, dy/dx = −1/2 1 − x, 1 − y 2 = x. Thus, d dy/dx = sin−1 y = dx 2 1 − y2 x(1 − x) −1

5A-4 a) y ′ = 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 difference: F (a) = a sinh(a) − cosh(a) Newton’s method for finding F (a) = 0, is the iteration an+1 = an − F (an )/F ′...

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