. Differential and integral calculus, an introductory course for colleges and engineering schools. and here we have x, y, and - ex- x pressed as trigonometric functions of the double area 2 u. Following the method presently to be employed with the hyperbolic functions, we shall derive these very simple relations again by means of definite integrals. We have u = area OAPB — ^ OAPB = H x dy = / Vl - y2 dy Jy=0 JO = - (yVl - y2 + sin1^) = -xy + ^sin-^. Hence 1 . sin-1 y and y = sm2u. Further, = VI x = v i — y2 = Vl — sin2 2 u = cos 2 u, and we have the same expressions for x and y as befo
. Differential and integral calculus, an introductory course for colleges and engineering schools. and here we have x, y, and - ex- x pressed as trigonometric functions of the double area 2 u. Following the method presently to be employed with the hyperbolic functions, we shall derive these very simple relations again by means of definite integrals. We have u = area OAPB — ^ OAPB = H x dy = / Vl - y2 dy Jy=0 JO = - (yVl - y2 + sin1^) = -xy + ^sin-^. Hence 1 . sin-1 y and y = sm2u. Further, = VI x = v i — y2 = Vl — sin2 2 u = cos 2 u, and we have the same expressions for x and y as before, but wehave obtained them here without making any use of the repre-sentation of 6 as an angle. Consider now the equilateral hyperbola x2 — y2 = 1, and let ube the area of the shaded sector. Then u = area OAPB — ~ OAPB = f^x dy = f Vl + y2 dy Jy=0 JO = \ [2/Vl + y2 + log (y + VF+y2)]= ^ xy + ^ log (y + vT+l/2). §255Therefore THE HYPERBOLIC FUNCTIONS u = g log (y + Vl + i/2) = - sh1 y, 397 P(»,y). and Further, We have then y = sh 2 m. a; = VT+Y2 = Vl + sh2 2 w = ch 2 w. ^ = ch 2 w, ?/ = sh 21£, - = th 2 w, which express the # and ?/ of a point on the equilateral hyperbolaas hyperbolic functions of the double area 2 u. We see thenthat 6 is twice the area u. Problem. Show, by the same method, that when the equation of thecurve is x2 — y2 = a2, 2u 2u x = a ch^ , y = ash—- CHAPTER XXXIIINTEGRATION OF RATIONAL FRACTIONS 256. Properties of Polynomials. We shall state a few of theproperties of polynomials that we shall have occasion to use, re-ferring the reader to algebra for the proofs of these properties. Let (1) P(x) = a0xn + ciix71-1-\-a2xn-2 + . . + an-2x2 + an-1x + an be a polynomial of the nth degree in x, the coefficients, a0, ah . .an being real constants. Such a polynomial has n roots. Letfi, r2, . . rn be these roots. Then it is a principle of algebrathat P(x) can be resolved into a product of n linear factors of the formx — r; that is, (2) P(x)
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