Unformatted text preview: es 41–43, verify the given integration formula. In each case, take a > 0. 41. 42. √ √ 1 a2 + x2 1 x 2 − a2 dx = sinh−1 dx = cosh
−1 x + C. a x + C. a is a solution of the equation which satisﬁes v(0) = 0. (b) Find lim v(t ).
t →∞ This limit is called the terminal velocity of the body. c 45. Use a CAS to ﬁnd a one-term expression for f .
(a) (b) f (x) dx = − f (x) dx = sech x(2 + tanh x) . 3(1 + tanh x)2 1 tanh−1 x + C If |x| < a. a 1 a 43. dx = 1 a2 − x 2 coth−1 x + C If |x| > a. a a 44. If a body of mass m falling from rest under the action of gravity encounters air resistance that is proportional to the square of its velocity, then the velocity v(t ) of the body at time t satisﬁes the equation m dv = mg − kv2 dt ln (x4 − 1) x2 tanh−1 x2 + . 2 4 Verify your results by differentiating the right-hand sides. c 46. Use a CAS to ﬁnd a one-term expression for f . x x (a) f (x) dx = 16 + x2 − 8 sinh−1 . 2 4
x2 + 1 x2 −1 x2 cosh−1 x2 + . 2 x2 + 1 2 Verify your results by differentiating the right-hand sides. (b) f (x) dx = − where k > 0 is the constant of proportionality and g is the gravitational constant. CHAPTER HIGHLIGHTS
View Full Document