258 Mr . Babbage’s new methods of investigating the 
this may be multiplied by and integrated any number of 
times in the same manner as (3), and the results would be 
2 
+ c+&a+c = 
1 . 2 ... 2 k 
-I 
X + X 
i ai sinl 
1 , 2 . . 2 k 2 
J 3 +ic 3 
3 2i sin 38 
x 5 -f .r 
zk — .1 
■&C. 
20 
q°g--) 2 * +l . (iog^) 2 *- 
1 . 2 . . 2^+1 ' 
—I 
X X 
zk + I • 
I sin 
1.2. .2k — 1 
x 3 — x 
zk+ 1 . „ 
3 ^ sin 39 
5 U sin 50' 
c + &c. + 
1 1 
i xi — X o 
+ -TTTT &C. 
(h 1 ) 
C 
2 k- 
2& + I . n 
5 sin 5 0 
"f" s l sin 50 &C. |(l,2) 
and these constants may be determined one from the other 
in the same manner as the former. I shall only give the 
value of the first, in order to compare the value of the series 
to which it is equal, with the sum of the same series deduced 
in another manner. 
710 , 2 ^ ±1 __ f I I 
L 1 2 ' 0 2 i+ i 3 ^ | i 2 sin 0 3* sin 30 1 5* sin 50 
In order to ascertain, the sums of series which contain cosines 
in their denominators, we must use an artifice which I shall 
now explain. 
Assuming as before -tyx = Ax -f- Ax 2 -f- Ar 3 -|- & c, and put- 
I 2 3 
ting v 2z for x, we have 
ypv 2z = Av Zz -{- Av 4z + Av 6z -f- &c. 
1 2 3 
If we were now to integrate this, we should introduce into 
the denominator of any term v il — l ; but we want to introduce 
the same expression with the signs of both terms positive. If 
we multiply both sides by ( — i) a and then integrate, we shall 
have first 
(— l) Av 2x (—1) + i) K + Av tz (—i) + &c. 
1 2 3 
