Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PRELIMINARY CALCULUS

πa 2 πa

2 a

x

y

Figure 2.14 The solution to exercise 2.17.

2.21 (a) 2(2−9cos^2 x)sinx;(b)(2x−^3 − 3 x−^1 )sinx−(3x−^2 +lnx)cosx;(c)8(4x^3 +
30 x^2 +62x+ 38) exp 2x.
2.23 (a) f(1) = 0 whilstf′(1)=0,andsof(x)mustbenegativeinsomeregionwith
x= 1 as an endpoint.
(b)f′(x)=tan^2 x>0andf(0) = 0;g′(x)=(−cosx)(tanx−x)/x^2 ,which is
never positive in the range.
2.25 The false result arises because tannxis not differentiable atx=π/(2n), which
lies in the range 0<x<π/n, and so the conditions for applying Rolle’s theorem
are not satisfied.
2.27 The relationship isxdy/dx=(2−x)y.
2.29 By implicit differentiation,y′(x)=(3x^2 −12)/(8− 3 y^2 ), givingy′(±2) = 0. Since
y(2) = 4 andy(−2) = 0, the curve touches thex-axis at the point (− 2 ,0).
2.31 (a) Express in partial fractions;J=^13 ln[(x−1)^4 /(x+2)]+c.
(b) Divide the numerator by the denominator and express the remainder in
partial fractions;J=x^2 /4+4ln(x+2)−3ln(x+3)+c.
(c) After division of the numerator by the denominator, the remainder can be
expressed as 2(x+3)−^1 −5(x+3)−^2 ;J=3x+2ln(x+3)+5(x+3)−^1 +c.
(d) Setx^4 =u;J=(4a^4 )−^1 tan−^1 (x^4 /a^4 )+c.

2.33 Writingb^2 − 4 acas ∆^2 >0, or 4ac−b^2 as ∆′^2 >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 thatdx=2(b−a)cosθsinθdθ.
(a)π;(b)π(b−a)^2 /8; (c)π(b−a)/2.
2.39 (a) (2−y^2 )cosy+2ysiny−2; (b) [(y^2 lny)/2] + [(1−y^2 )/4];
(c)ysin−^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 ofIn.
2.45 J 2 r+1=rJ 2 r− 1 and 2J 2 r=(2r−1)J 2 r− 2.
2.47 Setη=1−(x/a)^2 throughout, andx=asinθin one of the bounds.

2.49 L=

∫x

0

(

1+^14 exp 2x

) 1 / 2

dx.