Download PDF

integration of rational function of sine
and cosine∗
pahio†
2013-03-21 22:52:49
The integration task
Z
R(sin x, cos x) dx,
(1)
where the integrand is a rational function of sin x and cos x, changes via the
Weierstrass substitution
x
tan
= t
(2)
2
to a form having an integrand that is a rational function of t. Namely, since
x = 2 arctan t, we have
dx = 2 ·
1
dt,
1+t2
2t
,
1+t2
cos x =
(3)
and we can substitute
sin x =
1−t2
,
1+t2
(4)
getting
Z
Z
R(sin x, cos x) dx = 2
R
2t
1−t2
,
2
1+t 1+t2
dt
.
1+t2
Proof of the formulae (4): Using the double angle formulas of sine and cosine
and then dividing the numerators and the denominators by cos2 x2 we obtain
sin x =
2 sin x2 cos x2
2 tan x2
2t
=
2 x
2 x = 1 + t2 ,
x
2
1
+
tan
sin 2 + cos 2
2
cos x =
cos2 x2 − sin2
sin2 x2 + cos2
x
2
x
2
=
1 − tan2
1 + tan2
∗ hIntegrationOfRationalFunctionOfSineAndCosinei
x
2
x
2
=
1 − t2
.
1 + t2
created: h2013-03-21i by: hpahioi
version: h39380i Privacy setting: h1i hTopici h26A36i
† This text is available under the Creative Commons Attribution/Share-Alike License 3.0.
You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
1
Z
Example. The above formulae give from
Z
dx
=
sin x
Z
1+t2
1
dt =
·2·
2t
1+t2
Z
dx
the result
sin x
dt
x = ln |t| + C = ln tan + C
t
2
(which can also be expressed in the form − ln | csc x + cot x| + C; see the goniometric formulas).
Note 1. The substitution (2) is sometimes called the “universal trigonometric substitution”. In practice, it often gives rational functions that are too
complicated. In many cases, it is more profitable to use other substitutions:
R
• In the case R(sin x) cos x dx the substitution sin x = t is simpler.
R
• Similarly, in the case R(cos x) sin x dx the substitution cos x = t is
simpler.
• If the integrand depends only on tan x, the substitution tan x = t is
simpler.
• If the integrand is of the form R(sin2 x, cos2 x), one can use the substitution tan x = t; then
1
1
t2
dt
2
2
,
sin
x
=
1−cos
x
=
, dx =
.
cos2 x =
=
2
2
2
1
+
t
1
+
t
1
+
t2
1 + tan x
Z
dx
Example. The integration of
dx is of the last case:
cos4 x
Z
Z
Z
Z
dx
1
dt
t3
1
2 2
2
dx
=
dx
=
(1+t
)
·
=
(1+t
)
dt
=
+t+C = tan3 x+tan x+C.
4
2
2
2
cos x
(cos x)
1+t
3
3
Z
Z
dx
Example. The integral I =
dx = sec3 x dx is a peculiar case in
cos3 x
which one does not have to use the substitutions mentioned above, as integration
by parts is a simpler method for evaluating this integral. Thus,
u = sec x ⇒ du = sec x tan x dx;
dv = sec2 x dx ⇒ v = tan x.
Therefore,
Z
I
=
sec3 x dx
Z
= sec x tan x −
Z
sec x tan2 x dx
sec x (sec2 x − 1) dx
Z
= sec x tan x − I + sec x dx,
= sec x tan x −
2
and consequently
Z
dx
1
dx =
sec x tan x + ln | sec x + tan x| + C.
3
cos x
2
Note 2. There is also the “universal hyperbolic substitution” for integrating
rational functions of hyperbolic sine and cosine:
tanh
x
= t,
2
dx =
2dt
,
1−t2
sinh x =
2t
,
1−t2
cosh x =
1+t2
1−t2
References
[1] Л. Д. Кдрячев: Математичецкии анализ. Издательство “ВүсшаяШкола”. Москва (1970).
3