A Direct Proof of the Integral Formula for Arctangent Arnold J. Insel, Illinois State University, Normal, IL The College Mathematics Journal, May 1989, Volume 20, Number 3, pages 235–237. n this capsule, we give a direct proof that the Arctangent is an integral of 1ys1 1 x2d. It then becomes possible to use the Arctangent to determine the tangent and the other trigonometric functions. Here (Figure 1) for any real number a, we define Arctan a as the angle u (in radians) determined by angle OPR, where u is taken as negative if a < 0.
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In what follows, we fix a number a > 0. This will determine two regions, as shown in Figure 2. The region above the x-axis is bounded by the graph of y 5 1ys2s1 1 x2dd and the x-axis, where 0 ≤ x ≤ a. Therefore, the total area of this region is
E
a
0
dx . 2s1 1 x2d
Figure 1
Figure 2
The region below the x-axis is a sector of a circle having center s0, 21d and radius 1. The sides of the sector are determined by the y-axis and the line connecting s0, 21d to sa, 0d. Thus, these sides determine the angle with value Arctan a. Since the area of a sector of a circle of radius r and angle u (in radians) is r2uy2, the total area of this shaded region is 1y2 Arctan a. We shall show that these two shaded regions have equal areas. From this, it follows that the Arctangent can be represented as an integral of the function y 5 1ys1 1 x2d. First, consider the region above the x-axis (Figure 2). This region is divided into two subregions, A1 and A2. The rectangle A2 has area !sA2d 5 ays2s1 1 a2dd. The shaded sector below the x-axis is also divided into two subregions, B1 and B2. Since triangle CPD is similar to triangle OPR, the legs PC and CD of triangle CPD have lengths 1y!1 1 a2 and ay!1 1 a2, respectively.
Thus, B2 has area !sB2d 5 ays2s1 1 a2dd. In particular, !sA2d 5 !sB2d. It remains to be shown that !sA1d 5 !sB1d. First, solve the equation y 5 1y2s1 1 x2d for x to obtain x 5 !1 2 2yy!2y. Then integrate this along the y-axis to obtain
!sA1d 5
E
!1 2 2y
1y2
!2y
1y2s11a2d
dy.
Likewise, the circular boundary of B1 can be represented as the graph of x 5 !1 2 s y 1 1d2, where 1y!1 1 a2 2 1 ≤ y ≤ 0. Therefore,
!sB1d 5
E
0
s
1y!11a2
!1 2 s y 1 1d2 dy.
d21
Finally, we show that the integral for !sA1d can be transformed into the integral for !sB1d by means of the substitution t 5 !2y 2 1. Indeed, dt 5 dyy!2y and st 1 1d2 5 2y. Therefore,
!sA1d 5 5
E E
!1 2 2y
1y2
1y2s11a2d 0
s
1y!11a2
d 21
!2y
dy
!1 2 st 1 1d2 dt
5 !sB1d. We have therefore shown that
!sA1d 1 !sA2d 5 !sB1d 1 !sB2d. Thus,
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dx 1 5 Arctan a 2 2s1 1 x d 2
E
dx 5 Arctan a s1 1 x2d
a
0
or
a
0
(*)
for a > 0. A simple symmetry argument establishes the validity of (*) for a < 0. Equation (*) is clearly valid for a 5 0. Thus, (*) is valid for all real values a. We outline a method for obtaining the derivatives of the trigonometric functions from (*). First, apply the fundamental theorem of calculus to obtain the derivative of the Arctangent. The function f sxd 5 tan x s2 py2 < x < py2d is the inverse of the Arctangent, and its derivative f9sxd 5 sec2 x can be obtained from the inverse function theorem. Since the tangent function is a repetition of f on all intervals of the form ssn 2 s1y2ddp, sn 1 s1y2ddpd, we have d tan x 5 sec2 x. dx 2
Next, use the tangent function to represent the secant, and differentiate to obtain the usual formula for the derivative of the secant. For the derivatives of the sine and cosine, observe that cos x 5 1yssec xd and sin x 5 tan x cos x for x Þ sn 1 s1y2ddp. Differentiate to obtain the usual formulas with this restriction which can be removed by use of the identities.
1
sin x 5 cos x 2
p 2
2 and cos x 5 sin1x 1 p2 2.
Finally, the derivatives of the cotangent and cosecant can be obtained from the derivatives of the sine and cosine in the usual way.
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