Mathematical Methods for Physics and Engineering - Matematica.NET

PRELIMINARY CALCULUSy2aπa2πaxFigure 2.14 The solution to exercise 2.17.2.21 (a) 2(2 − 9cos 2 x)sinx; (b)(2x −3 − 3x −1 )sinx − (3x −2 +lnx)cosx; (c)8(4x 3 +30x 2 +62x + 38) exp 2x.2.23 (a) f(1) = 0 whilst f ′ (1) ≠0,andsof(x) mustbenegativeinsomeregionwithx = 1 as an endpoint.(b) f ′ (x) =tan 2 x>0andf(0) = 0; g ′ (x) =(− cos x)(tan x − x)/x 2 , which isnever positive in the range.2.25 The false result arises because tan nx is not differentiable at x = π/(2n), whichlies in the range 0 0:(i) ∆ −1 ln[(2ax + b − ∆)/(2ax + b +∆)]+k;(ii) 2∆ ′−1 tan −1 [(2ax + b)/∆ ′ ]+k;(iii) −2(2ax + b) −1 + k.2.35 f ′ (x)=(1+sinx)/ cos 2 x = f(x)secx; J =ln(f(x)) + c =ln(secx +tanx)+c.2.37 Note that dx =2(b − a)cosθ sin θdθ.(a) π; (b)π(b − a) 2 /8; (c) π(b − a)/2.2.39 (a) (2 − y 2 )cosy + 2y sin y − 2; (b) [(y 2 ln y)/2] + [(1 − y 2 )/4];(c) y sin −1 y +(1− y 2 ) 1/2 − 1;(d) ln(a 2 +1)− (1/y)ln(a 2 + y 2 )+(2/a)[tan −1 (y/a) − tan −1 (1/a)].2.41 Γ(n +1)=nΓ(n); (a) (i) n!, (ii) 15 √ π/8; (b) −2 √ π.2.43 By integrating twice, recover a multiple of I n .2.45 J 2r+1 = rJ 2r−1 and 2J 2r =(2r − 1)J 2r−2 .2.47 Set η =1− (x/a) 2 throughout, and x = a sin θ in one of the bounds.2.49 L = ∫ x0(1+14 exp 2x) 1/2dx.82