Article 26
Elementary Integrals.

The following useful indefinite integrals follow from Arts. 14, 15, 23:

Hyperbolic.
Circular.
1.sinhudu = coshu, sinudu = cosu,
2.coshudu = sinhu, cosudu = sinu,
3.tanhudu = logcoshu, tanudu = logcosu,
4.cothudu = logsinhu, cotudu = logsinu,
5.cschudu = logtanh u 2, cscudu = logtan u 2,
= sinh1(cschu), = cosh1(cschu),
6.sechudu = gdu, secudu = gd1u,
7. dx x2 + a2 = sinh1x a,4 dx a2 x2 = sin1x a,
8. dx x2 a2 = cosh1x a, dx a2 x2 = cos1x a,
9. dx a2 x2 x<a = 1 atanh1x a, dx a2 + x2 = 1 atan1x a,
10. dx x2 a2 x>a = 1 acoth1x a, dx a2 + x2 = 1 acot1x a,
11. dx xa2 x2 = 1 asech1x a, dx xx2 a2 = 1 asec1x a,
12. dx xa2 + x2 = 1 acsch1x a, dx xx2 a2 = 1 acsc1x a.

From these fundamental integrals the following may be derived:

13. dx ax2 + 2bx + c = 1 asinh1 ax + b ac b2,a  positive, ac > b2; = 1 acosh1 ax + b b2 ac,a  positive, ac < b2; = 1 acos1 ax + b b2 ac,a  negative.
14. dx ax2 + 2bx + c = 1 ac b2 tan1 ax + b ac b2,ac > b2; = 1 b2 actanh1 ax + b b2 ac,ac < b2,ax + b < b2 ac; = 1 b2 accoth1 ax + b b2 ac,ac < b2,ax + b > b2 ac;

Thus,

45 dx x2 4x + 3 = coth1(x 2) 45 = coth12 coth13 = tanh1(.5) tanh1(.3333) = .5494 .3466 = .2028.5 22.5 dx x2 4x + 3 = tanh1(x 2) 22.5 = tanh10 tanh1(0.5) = .5494.

(By interpreting these two integrals as areas, show graphically that the first is positive, and the second negative.)

15. dx (a x)x b = 2 a btanh1x b a b,  or  2 b atan1x b b a,  or  2 a bcoth1x b a b; the real form to be taken. (Put x b = z2, and apply 9, 10.)

16. dx (a x)b x = 2 b atanh1b x b a,  or  2 b acoth1b x b a,  or  2 a btan1b x a b; the real form to be taken.
17.(x2 a2)1 2 dx = 1 2x(x2 a2)1 2 1 2a2 cosh1x a.

By means of a reduction-formula this integral is easily made to depend on 8. It may also be obtained by transforming the expression into hyperbolic functions by the assumption x = acoshu, when the integral takes the form

a2sinh2udu = a2 2 (cosh2u 1)du = 1 4a2(sinh2u 2u) = 1 2a2(sinhucoshu u),

which gives 17 on replacing acoshu by x, and asinhu by (x2 a2)1 2 . The geometrical interpretation of the result is evident, as it expresses that the area of a rectangular-hyperbolic segment AMP is the difference between a triangle OMP and a sector OAP.

18.(a2 x2)1 2 dx = 1 2x(a2 x2)1 2 + 1 2a2 sin1x a. 19.(x2 + a2)1 2 dx = 1 2x(x2 a2)1 2 + 1 2a2 sinh1x a. 20.sec3φdφ =(1 + tan2φ)1 2 dtanφ = 1 2tanφ(1 + tan2φ)1 2 + 1 2sinh1(tanφ) = 1 2secφtanφ + 1 2gd1φ. 21.sech3udu = 1 2sechutanhu + 1 2gdu.
Prob. 71.
What is the geometrical interpretation of 18, 19?
Prob. 72.
Show that (ax2 + 2bx + c)1 2 dx reduces to 17, 18, 19, respectively: when a is positive, with ac < b2; when a is negative; and when a is positive, with ac > b2.
Prob. 73.
Prove sinh utanh udu = sinh u gd u, cosh ucoth udu = cosh u + log tanh u 2 .
Prob. 74.
Integrate (x2 + 2x + 5)1 2 dx, (x2 + 2x + 5)1dx, (x2 + 2x + 5)1 2 dx.
Prob. 75.
In the parabola y2 = 4px, if s be the length of arc measured from the vertex, and φ the angle which the tangent line makes with the vertical tangent, prove that the intrinsic equation of the curve is ds dφ = 2psec 3φ, s = psec φtan φ + pgd 1φ.
Prob. 76.
The polar equation of a parabola being r = asec 2θ, referred to its focus as pole, express s in terms of θ.
Prob. 77.
Find the intrinsic equation of the curve y a = cosh x a, and of the curve y a = log sec x a.
Prob. 78.
Investigate a formula of reduction for cosh nxdx; also integrate by parts
cosh 1xdx, tanh 1xdx, (sinh 1x)2dx; and show that the ordinary methods of reduction for cos mxsin nxdx can be applied to cosh mxsinh nxdx.