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Math 113
The Weierstrass Substitution
The Weierstrass substitution enables any rational function of the regular six trigonometric functions to
be integrated using the methods of partial fractions. It uses the substitution of
x
u = tan
.
(1)
2
The full method are substitutions for the values of dx, sin x, cos x, tan x, csc x, sec x, and cot x. Using
the identity tan2 θ + 1 = sec2 θ, the derivative of (1) is
x
x i
1
1h
1
du = sec2
dx =
1 + tan2
dx =
1 + u2 dx.
2
2
2
2
2
It follows that
dx =
To derive the substitutions for sin x and the
other trigonometric substitutions, refer to figure
1 and use the double angle identitities for sin x
and cos x. The double angle identity for sin x is
x
x
sin x = 2 sin
cos
2
2
and for cos x, the double angle identity is
x
x
cos x = cos2
− sin2
.
2
2
2 du
.
1 + u2
(2)
b
b
b
b
u
b √
b 1 + u2
b
b
b
b
b
b
x
2
b
b
b
1
Figure 1: Reference triangle for u = tan( x2 )
The substitution for sin x is
sin x = 2 sin
x
2
cos
x
2
u
1
2u
√
=2 √
=
2
2
1
+
u2
1+u
1+u
(3)
2 2
1
u
1 − u2
= √
− √
=
1 + u2
1 + u2
1 + u2
(4)
Similarly, for cos x, it is
cos x = cos2
x
2
− sin2
x
2
By using (3) and (4), the substitutions for tan x, csc x, sec x, and cot x is
2u
1+u2
1−u2
1+u2
tan x =
sin x
=
cos x
cot x =
1
1 − u2
=
tan x
2u
=
2u
1 − u2
csc x =
1 + u2
1
=
sin x
2u
sec x =
1
1 + u2
=
cos x
1 − u2
(5)
Note that the resulting equations for all 6 trigonometric functions, along with dx all are simple
polynomials in u. Hence, integrals of rational functions of trigonometric functions can be solved using
partial fractions. In summary,
u =
dx
=
x
tan
2
sin x =
2u
1 + u2
csc x =
1 + u2
2u
tan x
=
2u
1 − u2
2 du
1 + u2
cos x =
1 − u2
1 + u2
sec x =
1 + u2
1 − u2
cot x
=
1 − u2
2u
The Weierstrass Substitution, page 2
Example:
A short example can illustrate the power of the method. The integral of sec x is known to be
Z
sec x dx = ln | sec x + tan x| + C,
x+tan x
which is found by employing the “trick” of mutiplying the integrand by sec
sec x+tan x and employing the u
substitutiton u = sec x + tan x. A straightforward solution can be found using the Weierstrass method.
It follows that
Z
Z
1 + u2 2
dx
sec x dx =
1 − u2 1 + u2
Z
2
=
du
1 − u2
Z
1
1
=
+
du
1−u 1+u
= − ln |1 − u| + ln |1 + u| + C
1 + u
+C
= ln 1 − u
1 + tan x 2
= ln +C
1 − tan x2 Note that this does not look like ln | sec x + cos x|. However, using a bit of trickery (multiply by one)
always helps!
Z
1 + u
+C
sec x dx = ln 1 − u
1 + u 1 + u
+C
·
= ln 1 − u 1 + u
(1 + u)2 +C
= ln 1 − u2 1 + 2u + u2 +C
= ln 1 − u2 1 + u2
2u = ln +
+C
1 − u2
1 − u2 = ln |sec x + tan x| + C
Hence, the two solutions are identical!