Integration of sec θ, csc θ, sec3 θ and csc3 θ
These notes show several ways to integrate sec θ, csc θ, sec3 θ and csc3 θ.
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sec θ dθ — by trickery
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sec θ dθ — by partial fractions
sec θ+tan θ
The standard trick used to integrate sec θ is to multiply the integrand by 1 = sec
and
θ+tan θ
2
then substitute y = sec θ + tan θ, dy = (sec θ tan θ + sec θ) dθ.
Z
Z
Z
Z
sec2 θ + sec θ tan θ
dy
sec θ + tan θ
dθ =
dθ =
= ln |y| + C
sec θ dθ = sec θ
sec θ + tan θ
sec θ + tan θ
y
= ln | sec θ + tan θ| + C
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Another method for integrating sec θ dθ, that is more tedious, but less dependent on a
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memorized trick, is to convert sec θ dθ into the integral of a rational function using the
substitution y = sin θ, dy = cos θ dθ and then use partial fractions.
Z
Z
Z
Z
Z
cos θ
cos θ
1
1
dθ =
dθ =
dy
sec θ dθ =
dθ =
2
2
cos θ
cos θ
1 − y2
1 − sin θ
Z h
Z
1
1
1 h 1
1 i
1 i
=
dy =
dy
+
−
2 1+y 1−y
2
y+1 y−1
1
1 y + 1 =
ln |y + 1| − ln |y − 1| + C = ln +C
2
2
y−1
1 sin θ + 1 = ln +C
2
sin θ − 1
To see that this answer is really the same as the one above, note that
=⇒
sin θ + 1
=
sin θ − 1
1 sin θ + 1 ln =
2
sin θ − 1
(sin θ + 1)2
(sin θ + 1)2
=
− cos2 θ
sin2 θ − 1
sin θ + 1 1 (sin θ + 1)2 1 (sin θ + 1)2 ln = ln = ln 2
2
2
− cos θ
2
cos θ
cos θ
= ln | tan θ + sec θ|
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csc θ dθ — by the x = tan 2θ substitution
The integral
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csc θ dθ may also be evaluated by both the methods above. The answers are
Z
1 cos θ + 1 csc θ dθ = ln | cot θ − csc θ| + C = − ln +C
2
cos θ − 1
c Joel Feldman. 2014. All rights reserved.
1
February 1, 2014
Since csc θ is a rational function of sin θ and cos θ, the substitution
x = tan
θ
2
θ = 2 tan−1 x
dθ =
2
dx
1 + x2
θ
θ
x
1
2x
√
sin θ = 2 sin cos = 2 √
=
2
2
2
2
1 + x2
1+x
1+x
cos θ = cos2
converts
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√
1 + x2
θ
θ
1
x2
1 − x2
− sin2 =
−
=
2
2
1 + x2 1 + x2
1 + x2
x
θ/2
1
csc θ dθ into the integral of a rational function.
Z
Z
Z
Z
1 + x2
1
2
1
dθ =
dx =
dx = ln |x| + C
csc θ dθ =
2
sin θ
2x 1 + x
x
θ = ln tan + C
2
To see that this answer is really the same as the one above, note that
−2 sin2 (θ/2)
θ
cos θ − 1
=
= − tan
cot θ − csc θ =
sin θ
2 sin(θ/2) cos(θ/2)
2
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(1)
sec3 x dx — by trickery
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The standard trick used to evaluate sec3 x dx is integration by parts with u = sec x,
dv = sec2 x dx, du = sec x tan x dx, v = tan x.
Z
Z
Z
3
2
sec x dx = sec x sec x dx = sec x tan x − tan x sec x tan x dx
Since tan2 x + 1 = sec2 x, we have tan2 x = sec2 x − 1 and
Z
Z
Z
3
3
sec x dx = sec x tan x− [sec x−sec x] dx = sec x tan x+ln | sec x+tan x|+C− sec3 x dx
R
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where we used sec x dx = ln | sec x + tan x| + C. Now moving the sec3 x dx from the right
hand side to the left hand side
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2 sec3 x dx = sec x tan x + ln | sec x + tan x| + C
Z
1
1
⇒ sec3 x dx = sec x tan x + ln | sec x + tan x| + C
2
2
for a new arbitrary constant C.
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sec3 x dx — by partial fractions
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Another method for integrating sec3 x dx, that is more tedious, but less dependent on
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trickery, is to convert sec3 x dx into the integral of a rational function using the substitution
c Joel Feldman. 2014. All rights reserved.
2
February 1, 2014
y = sin x, dy = cos x dx and then use partial fractions.
Z
Z
Z
Z
Z
cos x
cos x
1
1
3
dx =
dx =
dy
sec x dx =
2 dx =
3
4
2
cos x
cos x
[1 − y 2 ]2
[1 − sin x]
Z
Z h 1
1
1 i2
1
=
dy
−
dy =
2 y−1 y+1
[y 2 − 1]2
Z
i
1
2
1
1 h
dy
−
+
=
4
(y − 1)2 (y − 1)(y + 1) (y + 1)2
Z
i
1 h
1
1
1
1
=
dy
−
+
+
4
(y − 1)2 y − 1 y + 1 (y + 1)2
1
1 i
1h
−
+C
− ln |y − 1| + ln |y + 1| −
=
4
y−1
y+1
1 2y
1 y
1 y + 1 1 y + 1 =− 2
+
C
=
+ ln +
ln +C
4y −1 4
y−1
2 1 − y2 4
y−1
1 sin x + 1 1 sin x
+ ln =
+C
2 cos2 x 4
sin x − 1
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csc3 θ dθ — by the x = tan 2θ substitution
As another example of the
x = tan
θ
2
dθ =
2
dx
1 + x2
sin θ =
2x
1 + x2
cos θ =
1 − x2
1 + x2
substitution, we evaluate
Z
Z Z
3
1 + x2 2
1
1 + 2x2 + x4
csc θ dθ =
dx
=
dx
2x
1 + x2
4
x3
1 h x−2
x2 i
+C
=
+ 2 ln |x| +
4 −2
2
θi
θ
θ 1h
− cot2 + 4 ln tan + tan2
+C
=
8
2
2
2
Z
3
1
dθ =
sin3 θ
By the usual double angle formulae
tan2
cos2 2θ
sin4 θ2 − cos4 2θ
sin2 θ2
θ
θ
−
=
− cot2 =
2
2
cos2 θ2
sin2 2θ
sin2 2θ cos2 2θ
sin2 2θ − cos2 2θ
=
sin2 θ2 cos2 2θ
− cos θ
= 1 2
sin θ
4
c Joel Feldman. 2014. All rights reserved.
3
since sin2 θ2 + cos2
θ
2
=1
February 1, 2014
so we may also write
Z
1
1
csc3 θ dθ = − cot θ csc θ + ln | cot θ − csc θ| + C
2
2
by (1).
c Joel Feldman. 2014. All rights reserved.
4
February 1, 2014