Download Reverse Trig Substitution

Reverse Trig Substitution
Math 133
Stewart §7.3
R
Reducing to standard trig forms. To find an indefinite integral f (x) dx, we transform it by methods like Substitution and Integration by Parts until we reduce to an
integral we recognize from before, a “standard form”. In the previous section §7.2, we
were able to compute most integrals involving products of trig functions, so these are
now standard forms to work toward.
√
A common type of difficult integral involves forms like ±x2 ± 1. We convert such
forms into trigonometric integrals, which at first seems to complicate them. However, we
take careful advantage of the Pythagorean identities cos2 (θ) + sin2 (θ) = 1 and tan2 (θ) +
1 = sec2 (θ), so that the resulting
trig formulas simplify to doable integrals.
R√
2
Our first example is
1 − x dx. This seems simple enough, but neither Substitution nor Integration by Parts will simplify it. Instead, imagine that we obtained this
from a more complicated integral by a trig substitution x = sin(θ): the current variable
x is actually a function of a previous variable θ. The previous integral would be:
(
Zp
Zp
x = sin(θ)
2
1−x2 dx =
1− sin (θ) · cos(θ) dθ, where
dx = cos(θ) dθ.
p We did not choose this substitution at random: the trig form simplifies because
1− sin2 (θ) = cos(θ), and we obtain a standard form from §7.2:
Z
Z p
2
cos2 (θ) dθ = 21 θ − 14 sin(2θ) = 12 θ − 12 sin(θ) cos(θ) .
1− sin (θ) · cos(θ) dθ =
Finally, we substitute back in terms of x: θ = arcsin(x), sin(θ) = x, cos(θ) =
Zp
p
1−x2 dx = 12 arcsin(x) − 21 x 1−x2 .
√
1−x2 :∗
√
Let’s check the area of the unit circle, i.e. twice the area under the graph y = 1−x2 :
Z 1p
ix=1
h
p
2
2
= arcsin(1) − arcsin(−1) = π.
2
1−x dx = arcsin(x) − x 1−x
x=−1
−1
p
Integrals with ±x2 ±a2 . We choose a reverse trig substitution depending on the
signs in the expression, then use the corresponding Pythagorean identity to obtain a
standard trig form:
√
√
√
a2 −x2
x = a sin(θ)
dx = a cos(θ) dθ
p
a2 −a2 sin2 (θ) = a cos(θ)
a2 +x2
x = a tan(θ)
dx = a sec2 (θ) dθ
p
a2 +a2 tan2 (θ) = a sec(θ)
x2 −a2
x = a sec(θ)
dx = a tan(θ) sec(θ) dθ
p
a2 sec2 (θ)−a2 = a tan(θ)
Z
example:
∗
p
1
dx. Let x = 2 sin(θ), dx = 2 cos(θ) dθ, 4−(2 sin(θ))2 = 2 cos(θ):
4 − x2
Z
Z
1
1
√
dx =
· 2 cos(θ) dθ = θ = arcsin( 12 x) .
2
2 cos(θ)
4−x
√
See §6.6 for inverse trig functions.
We could do this more directly by manipulating the integrand to the known derivative
of arcsin(x) by the substitution u = 21 x:
Z
Z
Z
1
1
1
√
q
√
dx =
du = arcsin(u) = arcsin( 12 x).
· 21 dx =
2
2
1
4−x
1−u
1−( 2 x)2
Z
example:
q
1
dx. Let x = 32 tan(θ), dx = 23 sec2 (θ) dθ, 9( 23 tan(θ))2 +4 = 2 sec(θ):
9x2 +4
Z
Z
Z
1
1
2
2
1
√
dx =
· sec (θ) dθ = 3 sec(θ) dθ
2 sec(θ) 3
9x2 +4
√
= 13 lntan(θ) + sec(θ) = 13 ln 23 x + 12 9x2 +4 .
√
We can even write this in terms of the inverse hyperbolic sine from §6.7:
3 p9 2 1
ln
x+
x
+1
= 13 sinh−1 ( 32 x) .
3
2
4
Z√
example:
p
x2 −25
dx; x = 5 sec(θ), dx = 5 tan(θ) sec(θ) dθ, (5 sec(θ))2 −25 = 5 tan(θ):
x
Z√ 2
Z
Z
5 tan(θ)
x −25
dx =
· 5 tan(θ) sec(θ) dθ = 5 tan2 (θ) dθ
x
5 sec(θ)
Z
p
= 5 sec2 (θ)−1 dθ = 5 tan(θ) − 5θ =
x2 −25 − 5 arcsec( x5 ) .
√
√
Since we dislike arcsec (§6.6), we can also write this as: x2 −25 − 5 arctan( 51 x2 −25) .
Z
3
1
example:
dx; x = 2 sec(θ), dx = 2 tan(θ) sec(θ) dθ, ((2 sec(θ))2 −4) 2 = 8 tan3 (θ):
3/2
2
(x −4)
Z
Z
Z
1
1
1
· 2 tan(θ) sec(θ) dθ = 4 sin−2 (θ) cos(θ) dθ
dx =
8 tan3 (θ)
(x2 −4)3/2
1
x
x
= − 14 sin(θ)−1 = − 14 q 2
= − √
.
4 x2 −4
( 12 x)2 −1
Extra topic: Geometric substitution. Here is some esoteric knowledge for the hardcore students. Consider any integral in which trig functions are combined by the four
arithmetic operations, such as:
Z
cos2 (x) sin(x) − 2 tan(x) + 1
dθ.
sec3 (θ) + sin3 (θ) + 3 cos(θ) sin(θ) + 5
There is an amazing technique, the Tangent Half-Angle Substitution, which allows us to
reduce any such problem to the integral of a rational function (a quotient of polynomials),
which can then be done by Partial Fractions (see §7.4).
This substitution is motivated entirely by geometry. Recall that the basic trig functions are circular functions: the coordinates (x, y) = (cos(θ), sin(θ)) for θ ∈ [0, 2π]
trace the points on a unit circle. There is another way to trace this circle, the rational
parametrization: for any number t ∈ (−∞, ∞), draw the line L from the fixed point
(−1, 0) to the point (0, t) on the y-axis: this line cuts the circle in exactly one other
point (x, y). As (0, t) moves along the y-axis, the point (x, y) moves around the entire
circle, leaving out only the fixed point (−1, 0).
To compute the coordinates (x, y) corresponding to a given t, we write the slope of line
L using the two similar triangles between L and the x-axis:
y
t
slope =
=
=⇒
y = t(x+1).
1
x+1
The point (x, y) also satisfies the circle equation x2 + y 2 = 1. Substituting for y gives:
x2 + (t(x+1))2 = 1
=⇒
x2 + t2 x2 + 2t2 x + t2 − 1 = 0 .
Now, for any t, this equation is always satisfied by the fixed point with x = −1, so x+1
must be a factor of the last polynomial. Long division gives:

2

x= 1−t
1 + t2
x2 + t2 x2 + 2tx + t − 1 = (x+1)(t2 x+x+t2 −1) = 0 =⇒

 or x = −1 .
Plugging this into y = t(x+1) to write y in terms of t, we get a new formula for the
points of the circle, controlled by t ∈ (−∞, ∞):
1 − t2
2t
.
(x, y) =
,
1 + t2 1 + t2
Comparing this with our trig formula (x, y) = (cos(θ), sin(θ)) suggests the substitution:
1 − t2
,
1 + t2
2t
,
etc.
1 − t2
2t
We can think of this as a backward substitution, θ = arcsin 1+t
2 , and compute:
cos(θ) =
sin(θ) =
=⇒
dθ = 2
sin(θ) =
2t
1 + t2
2t
,
1 + t2
=⇒
tan(θ) =
cos(θ) dθ = 2
1 − t2
dt
(1 + t2 )2
1 − t2
1 + t2 1 − t2
2
1
dt
=
2
dt =
dt .
2
2
2
2
2
cos(θ) (1 + t )
1 − t (1 + t )
1 + t2
To restore the original variable θ at the end of the integration, we need to write
y
t = x+1
in terms of trig functions. We can do this just by (x, y) = (cos(θ), sin(θ)); but
also recall the theorem of elementary geometry which says the angle between L and the
x-axis is 2θ , giving another expression for the slope:
y
sin(θ)
θ
slope = t =
=
= tan
,
x+1
cos(θ) + 1
2
which is why we call it the Tangent Half-Angle Substitution.
example: We carry out this substitution, then the Partial Fraction Method from §7.4:
Z
Z
Z
2
1
2(t2 +1)
1
·
dθ
=
dt
=
dt
2
2
1+t2
t4 + 8t2 + 3
sin2 (θ) + cos(θ) + 2
2t
+ 1−t + 2
1+t2
1+t2
Z 1 + √3
1 − √313
2(t2 +1)
13
√
√
dt =
=
+
dt
(t2 +4)2 − 13
t2 +4+ 13
t2 +4− 13
p √
p √
√
√
t
√
= ( 13+3) 4− 13 arctan
+ ( 13−3) 4+ 13 arctan √
√
Z
4+ 13
t
√
4− 13
p √
p √
√
√
tan(θ/2)
tan(θ/2)
= ( 13+3) 4− 13 arctan √ √
+ ( 13−3) 4+ 13 arctan √ √
.
4+ 13
4− 13
The point here is not the specific answer, which can be gotten by computer much
more reliably than by hand. It is the principle that this is possible for any integral of this
type, precisely because of the rational parametrization of the circle. This leads toward
the theory of Lie groups, which generalizes the circle to highly symmetric geometric
objects in higher dimensions, starting with the 3-dimensional sphere.†
†
We should be able to produce a similar substitution for integrals involving matrix coefficients of any
representation of a compact Lie group, composed with the exponential map on the Lie algebra.