Integration of sec x and sec 3 x
R sec x dx – by trickery
The standard trick used to integrate sec x is to multiply the integrand by 1 = sec x +tan x sec x +tan x then substitute y = sec x + tan x , dy = (sec x tan x + sec 2 x ) dx .
and
Z sec x dx =
Z sec x sec x +tan x sec x +tan x dx =
Z sec
2 x +sec x tan x sec x +tan x dx =
Z dy y
= ln | y | + C
= ln | sec x + tan x | + C
R sec x dx – by partial fractions
Another method for integrating R sec x dx , that is more tedious, but less dependent on a memorized trick, is to convert R sec x dx into the integral of a rational function using the substitution y = sin x , dy = cos x dx and then use partial fractions.
Z Z Z Z Z sec x dx = 1 cos x dx = cos x cos 2 x dx =
1 cos x
− sin 2 x dx = 1
1
− y 2 dy
=
Z
1
2 h
1
1+ y
+ 1
1
− y i dy = 1
2
Z h
1 y +1
− 1 y
−
1 i dy = 1
2 h ln | y + 1 | − ln | y − 1 | i
+ C
= 1
2 ln y +1 y
−
1
+ C = 1
2 ln sin x +1 sin x
−
1
+ C
To see that this answer is really the same as the one above, note that
⇒ 1
2 sin x +1 sin x
−
1 ln
=
(sin x +1)
2 sin 2 x
−
1 sin x +1 sin x
−
1
= 1
2 ln
=
(sin x +1)
2
− cos 2 x
(sin x +1)
2
− cos 2 x
= 1
2 ln
(sin x +1)
2 cos 2 x
= ln sin x +1 cos x
= ln | tan x + sec x |
R sec 3 x dx – by trickery
The standard trick used to evaluate R sec 3 x dx is integration by parts with u = sec x , dv = sec 2 x dx , du = sec x tan x dx , v = tan x .
Z sec
3 x dx =
Z sec x sec
2 x dx = sec x tan x −
Z tan x sec x tan x dx
Since tan 2 x + 1 = sec 2 x , we have tan 2 x = sec 2 x − 1 and
Z sec
3 x dx = sec x tan x −
Z
[sec
3 x − sec x ] dx = sec x tan x +ln | sec x +tan x | + C −
Z sec
3 x dx where we used R sec x dx = ln | sec x + tan x | + C . Now moving the R sec 3 x dx from the right hand side to the left hand side
Z
2 sec
3 x dx = sec x tan x + ln | sec x + tan x | + C
⇒
Z sec
3 x dx = 1
2 sec x tan x + 1
2 ln | sec x + tan x | + C for a new arbitrary constant C .
c Joel Feldman. 2001. All rights reserved.
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R sec 3 x dx – by partial fractions
Another method for integrating trickery, is to convert R
R sec 3 x dx , that is more tedious, but less dependent on sec 3 x dx into the integral of a rational function using the substitution y = sin x , dy = cos x dx and then use partial fractions.
Z sec 3 x dx =
Z
=
Z h
1 cos 3 x
1
2 dx =
Z cos x cos 4 x dx =
Z
[1
− cos x sin 2 x ] 2 dx =
Z
[1
−
1 y 2 ] 2 dy =
Z
1 y
−
1
− 1 y +1 i
2 dy = 1
4
Z h
( y
1
−
1) 2
−
( y
−
2
1)( y +1)
+ 1
( y +1) 2 i dy
[ y 2
1
−
1] 2 dy
= 1
4
Z h
( y
1
−
1) 2
− y
1
−
1
+ 1 y +1
+ 1
( y +1) 2 i dy
=
=
1
4 h
−
1
2 y
1
− y 2 y
1
−
1
+ 1
4
− ln | y − 1 | + ln | y + 1 | − 1 y +1 i
+ C = − 1
4 y
2 y
2
−
1 ln y +1 y
−
1
+ C = 1
2 sin x cos 2 x
+ 1
2 ln sin x +1 sin x
−
1
+ C
+ 1
4 ln y +1 y
−
1
+ C c Joel Feldman. 2001. All rights reserved.
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