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then the tangential equation of its involute is 
j'-i sin <p = J sin <p d <\ 5 -f- 3 J cos’ <p sin 1 <p 
+ 3 j*cos s <J3 sin’ <p (ftp+Jcos <P sin <P (135) 
Of these four integrals, the first and fourth are elementary, and the third is derived 
from the second by putting (j> — in place of (p and changing signs. Hence the 
question will be solved if we integrate 
J cos’ <p sin 1 d<p. 
To reduce this to elliptic integrals, let z 3 = cot 3 <p, and we easily find 
f i - • • - sin^ <p . C dz 
jcos.<p S in T <(l= c ^ + j ?VT7; ,. 
Now 
VTT^)_i * 1 
dz 
Hence 
z ) 2 * n/ 1+5 3 V i+i| 
2 dz 
Jco8‘4>sin» 4 irf 1 i=-^+ij , 
(136) 
and the question is completely solved. (See art. 32.) 
(4) Let *=:« cos <p be the tangential equation of a curve, then (see art. 50) the intrinsic 
equation of its involute is 
, ={ i +( 4 ) , }^) s!n ‘* 1 
3 a sin 2tp a<p 
: b rT 
. (137) 
(5) Let v—a cos <p ; then we find for the involute 
i _2a sin 3 <p , a<p 
9 "'"y 
(133) 
(6) Let the intrinsic equation of a curve be 
s=a cos 3 <J5 ; 
then (see art. 51) the tangential equation of its involute is 
a C cos 4 cpdcp 
4J sin 2 <p 
=~ 6 (4 cot <p+2(p-hsin 2<p) . (139) 
Similarly, if s=acos 5 <p, 
v=~ { 16 cot <p -f- 30 <p-f-9 sin 2<p — 4 sin 3 <p cos <p \. . . . (140) 
is its involute. 
