Difference Equations to Differential Equations — Section 6.6: Trigonometric Substitutions
Section 6.6
Difference Equations
to
Differential Equations
In the last section we saw that
Z
Trigonometric Substitutions
√
1
dx = sin−1 (x) + c.
1 − x2
However, we arrived at this result as a consequence of our differentiation of the arc sine
function, not as the outcome of the application of some systematic approach to the evaluation of integrals of this type. In this section we will explore how substitutions based on the
arc sine, arc tangent, and arc secant functions provide a systematic method for evaluating
integrals similar to this one.
Sine substitutions
To begin, consider evaluating
Z
√
1
dx
1 − x2
by using the substitution u = sin−1 (x). The motivation for such a substitution stems from
the fact that, for − π2 ≤ u ≤ π2 , u = sin−1 (x) if and only if x = sin(u). In the latter form,
we see that
dx = cos(u)du
and
q
p
p
2
1 − x = 1 − sin2 (u) = cos2 (u) = | cos(u)|.
(6.6.1)
Since cos(u) ≥ 0 when π2 ≤ u ≤ π2 , (6.6.1) becomes
p
1 − x2 = cos(u).
Thus
Z
1
√
dx =
1 − x2
Z
cos(u)
du =
cos(u)
Z
du = u + c = sin−1 (x) + c.
Of course, there is nothing new in the result itself; it is the technique, which we may
generalize to other integrals of a similar type, which is of interest. Specifically, for an
integral with a factor of the form
p
a2 − x2
or
√
1
,
a2 − x2
1
Copyright c by Dan Sloughter 2000
2
Trigonometric Substitutions
Section 6.6
where a > 0, the substitution
x = a sin(u),
π
π
≤u≤ ,
2
2
may prove to be useful because of the simplification
q
q
p
p
2
2
2
2
2
a − x = a − a sin (x) = a2 (1 − sin2 (u)) = a cos2 (u) = a cos(u).
(6.6.2)
(6.6.3)
Although this substitution is equivalent to the substitution u = sin−1 xa , we will see that
it is more convenient to work with it in the form x = a sin(u).
Z
1
√
Example To evaluate the integral
dx, we make the substitution
9 − x2
π
π
<u< ,
2
2
dx = 3 cos(u)du.
x = 3 sin(u), −
Note that we omit both u = − π2 and u = π2 since the function being integrated is not
defined at either x = −3 or x = 3. Then
Z
Z
3 cos(u)
1
√
q
du
dx =
2
9 − x2
9 − 9 sin (u)
Z
3 cos(u)
p
=
du
3 1 − sin2 (u)
Z
cos(u)
p
=
du
cos2 (u)
Z
cos(u)
du
=
cos(u)
Z
= du
= u + c.
Now x = 3 sin(u) implies that u = sin−1 x3 , so we have
Z
x
1
√
dx = sin−1
+ c.
3
9 − x2
Example
Z p
To evaluate
4 − x2 dx, we make the substitution
π
π
≤u≤ ,
2
2
dx = 2 cos(u)du.
x = 2 sin(u), −
Section 6.6
Trigonometric Substitutions
3
2
x
u
4 x2
Figure 6.6.1 Right triangle with sin(u) = x2
Then
Z p
Z q
2
4 − x dx = 2
4 − 4 sin2 (u) cos(u)du
Z q
=4
1 − sin2 (u) cos(u)du
Z p
cos2 (u) cos(u)du
=4
Z
= 4 cos2 (u)du
Z
1 + cos(2u)
du
=4
2
Z
= 2 (1 + cos(2u))du
= 2u + sin(2u) + c
= 2u + 2 sin(u) cos(u) + c.
Since x = 2 sin(u), sin(u) = x2 and u = sin−1 x2 . Moreover,
cos2 (u) = 1 − sin2 (u) = 1 −
x2
4 − x2
=
,
4
4
so
1p
4 − x2 ,
2
where we have, once again, used the fact that cos(u) ≥ 0 since − π2 ≤ u ≤ π2 . Note that
this expression for cos(u) may also be deduced from Figure 6.6.1, where we have a right
triangle with an acute angle of size u such that sin(u) = x2 . Putting everything together,
we have
Z p
1 p
−1 x
2
4 − x dx = 2 sin
+ x 4 − x2 + c.
2
2
cos(u) =
Notice that a considerable amount of the work in the previous example involved expressing the answer in terms of x once it had been found in terms of u. The next example
4
Trigonometric Substitutions
Section 6.6
illustrates how this work is unnecessary when evaluating definite integrals since we can
change the limits of integration and, from that point on, do all our work in terms of u.
√
Example Recall that, for r > 0, the graph of y = r2 − x2 is the upper half of a circle
of radius r centered at the origin. Hence we should have
Z r p
r2 − x2 dr = πr2 .
2
−r
We are now able to verify this. Let
π
π
≤u≤ ,
2
2
dx = r cos(u)du.
x = r sin(u), −
Then u = sin−1
x
r
, so when x = −r,
π
u = sin−1 (−1) = − ,
2
and when x = r,
u = sin−1 (1) =
Thus
π
.
2
Z r p
Z π2 q
2
r2 − x2 dx = 2
r2 − r2 sin2 (u) r cos(u)du
−π
2
−r
Z π2
=2
r2
p
cos2 (u) cos(u)du
−π
2
= 2r
= 2r
=r
2
2
2
Z π2
cos2 (u)du
−π
2
π
2
Z
1 + cos(2u)
du
2
−π
2
Z π2
(1 + cos(2u))du
−π
2
sin(2u) π2
=r u+
2
−π
π
π 2
= r2
+0 − − +0
2
2
= πr2 .
2
Tangent substitutions
In a similar fashion, the substitution
x = a tan(u), −
π
π
<u< ,
2
2
(6.6.4)
Section 6.6
Trigonometric Substitutions
5
may be useful for integrals which have a factor of the form
p
a2 + x2 ,
√
1
,
a2 + x2
or
1
a2 + x2
because of the simplification
a2 + x2 = a2 + a2 tan2 (u) = a2 (1 + tan2 (u)) = a2 sec2 (u).
(6.6.5)
Note that with
our restriction on u, this substitution is equivalent to the substitution
u = tan−1 xa .
Z
1
Example To evaluate the integral
dx, we make the substitution
4 + x2
π
π
<u< ,
2
2
2
dx = 2 sec (u)du.
x = 2 tan(u), −
Then
Z
Since x = 2 tan(u), u = tan−1
Z
Z
Example
To evaluate
√
2sec2 (u)
du
4 + 4 tan2 (u)
Z
1
sec2 (u)
=
du
2
1 + tan2 (u)
Z
1
sec2 (u)
=
du
2
sec2 (u)
Z
1
du
=
2
1
= u + c.
2
1
dx =
4 + x2
x
2
Z
. Thus
1
1
−1 x
dx
=
tan
+ c.
4 + x2
2
2
1
dx, we make the substitution
1 + x2
π
π
<u< ,
2
2
2
dx = sec (u)du.
x = tan(u), −
6
Trigonometric Substitutions
Section 6.6
1 x2
x
u
1
Figure 6.6.2 Right triangle with tan(u) = x
Then
q
p
p
2
1 + x = 1 + tan2 (u) = sec2 (u) = | sec(u)|.
Since − π2 < u < π2 , sec(u) > 0, so | sec(u)| = sec(u). Hence
Z
1
dx =
1 + x2
Z
sec2 (u)
du
sec(u)
Z
=
sec(u)du
= log | sec(u) + tan(u)| + c,
where the final integral follows from an example in Section 6.2. Now tan(u) = x, so
sec2 (u) = 1 + tan2 (u) = 1 + x2 .
Since sec(u) > 0, it follows that
sec(u) =
p
1 + x2 .
Note that this expression for sec(u) may also be deduced from Figure 6.6.2, where we have
a right triangle with an acute angle of size u such that tan(u) = x. Thus
Z
p
1
√
dx = log
1 + x2 + x + c.
1 + x2
Secant substitutions
For integrals involving a factor of the form
p
x2 − a2
or
√
1
,
x2 − a2
Section 6.6
Trigonometric Substitutions
7
where a > 0, the substitution
x = a sec(u)
(6.6.6)
q
p
2
sec (u) − 1 = a tan2 (u) = a| tan(u)|.
(6.6.7)
may be useful. With this substitution,
p
x2 − a2 =
p
a2 sec2 (u) − a2 = a
√
Now x2 − a2 is meaningful only if either x ≥ a or x ≤ −a. Since x = a sec(u), the former
case corresponds to u in the interval [0, π2 ) and the latter to u in the interval ( π2 , π]. For
0 ≤ u < π2 , tan(u) ≥ 0, so
p
x2 − a2 = a tan(u);
for π2 < u ≤ π, tan(u) ≤ 0, so
p
x2 − a2 = −a tan(u).
Hence it is important when evaluating integrals of this type to be careful about which
values of x are of interest.
Z
1
√
Example To evaluate
dx for x > 3, we make the substitution
2
x −9
π
,
2
dx = 3 sec(u) tan(u)du.
x = 3 sec(u), 0 < u <
Then
Z
√
Z
1
x2 − 9
3 sec(u) tan(u)
p
du
9 sec2 (u) − 9
Z
3 sec(u) tan(u)
p
=
du
3 sec2 (u) − 1
Z
sec(u) tan(u)
p
=
du
tan2 (u)
Z
sec(u) tan(u)
=
du
tan(u)
Z
= sec(u)du
dx =
= log | sec(u) + tan(u)| + c.
Now sec(u) = x3 , so
tan2 (u) = sec2 (u) − 1 =
Hence
tan(u) =
x2
x2 − 9
−1=
.
9
9
1p 2
x − 9,
3
8
Trigonometric Substitutions
Section 6.6
x
x2
9
u
3
Figure 6.6.3 Right triangle with sec(u) = x3
where we have used the fact that tan(u) > 0 since 0 < u < π2 . Note that this expression
for sec(u) may also be deduced from Figure 6.6.3, where we have a right triangle with an
acute angle of size u such that sec(u) = x3 . Thus
Z
√
1
x2 − 9
dx = log
p
x 1p 2
+
x − 9 = log |x + x2 − 9| − log(3) + c.
3 3
Since log(3) is a constant, we may combine it with the arbitrary constant of integration.
Moreover, since we are assuming x > 3, we may remove the absolute value and write
Z
p
1
√
dx = log(x + x2 − 9) + c.
x2 − 9
Problems
1. Evaluate the following integrals.
Z
3
√
(a)
dx
16 − x2
Z p
5 − z 2 dz
(c)
Z
(e)
1
√
dx
4 + x2
2. Evaluate the following integrals.
Z p
(a)
z 1 − z 2 dz
Z
(c)
Z
(e)
√
1
x2 − 4
Z
(b)
Z
(d)
Z
(f)
(b)
4
√
dx
3 − 2x2
√
4x
dx
1 + x2
1
3
(1 − x2 ) 2
(d)
Z
(f)
x
dx
4 − x2
5
dx
6 + x2
Z
Z
dx, x > 2
√
√
1
x2 − 4
dx
dx, x < −2
3
dx
5 + 2x2
Section 6.6
Z p
4 − t2 dt
(g)
Trigonometric Substitutions
Z p
(h)
1 − 4x2 dx
3. Evaluate the following integrals.
Z 3p
(a)
9 − t2 dt
(b)
Z 1p
1 + x2 dx
−1
0
Z 1
(c)
0
1
Z 1
3 dx
(d)
(1 + x2 ) 2
x2
p
1 − x2 dx
0
Z 10
(e)
9
Z 2
1
√
dx
√
x2 − 25
5 2
(f)
4. Evaluate
Z
√
2
2 x
√
3
dx
x2 − 1
1
dx
1 − x2
using (a) partial fractions and (b) the substitution x = sin(u). How do the two methods
compare?
5. Evaluate
Z
√
1
dx
1 − x2
using the substitution x = cos(u) with 0 < u < π.
6. (a) Evaluate
Z
√
1
x2 − 9
dx
for x < −3.
(b) Show that
Z
√
1
x2 − 9
for both x > 3 and x < −3.
dx = log |x +
p
x2 − 9| + c,