Download Trigonometric Substitution Trig. Substitution is often used when the

Trigonometric Substitution
Trig. Substitution is often used when the integrand involves:
a2 − u 2
or
Throughout this handout,
a>0
a2 + u 2
u2 − a2 ,
or
is a positive constant.
SUMMARY CHART
then do
i.e. do
if f (u)
Goal: Find f (u) du
need
& get
trig.
sub.
trig. sub.
has
working
reality
↓recall↓
form
form
u
R du
−1
u
u
2
2
≤ 1 - π ≤ θ ≤ π (?1 )
√
=
sin
+
c
a
−
u
u
=
a
sin
θ
θ
=
arcsin
2
2
a
a
a
2
2
a −u
R du
u
π
π
1
−1 u
2
2
a + u u = a tan θ θ = arctan a
none
- 2 < θ < 2 (?2 )
= a tan a + c
a2 +u2
R
|u|
√ du
= a1 sec−1 a + c u2 − a2 u = a sec θ θ = arcsec ua
1 ≤ ua u u2 −a2
R
(1)
(2)
(3)
(3+ )
if u is positive
a≤u
(3− )
if u is negative
u≤-a
Helpful for when
(1)
(2)
√
π
2
(?3+ )
<θ≤π
(?3− )
0≤θ<
π
2
sign is involved.
If the integrand involves a2 − u2 , then let u = a sin θ and will have:
p
p
√
(4)
(?1 )
a2 − u2 =
a2 − a2 sin2 θ = a 1 − sin2 θ = a |cos θ| =
a cos θ
If the integrand involves a2 + u2 , then let u = a tan θ and will have:
√
√
√
(4)
(?2 )
a2 + u2 =
a2 + a2 tan2 θ = a 1 + tan2 θ = a |sec θ| =
a sec θ
(3+ )
If the integrand involves u2 − a2 , then let u = a sec θ and, in the case that a ≤ u, will have:
√
√
√
(?3+ )
(4)
+ a tan θ
u2 − a2 =
a2 sec2 θ − a2 = a sec2 θ − 1 = a |tan θ| =
(3− )
If the integrand involves u2 − a2 , then let u = a sec θ and, in the case that u ≤ −a, will have:
√
√
√
(?3− )
(4)
u2 − a2 =
a2 sec2 θ − a2 = a sec2 θ − 1 = a |tan θ| =
– a tan θ
(3)
If book does not provide enough info to determine which (3) to use, then use (3+ ).
Self-check : if (4) does not follow from cos2 θ + sin2 θ = 1, then you picked the wrong trig substitution.
Logic Reduces Memorization (take a = 1 here for simplicity)
(sin θ)2 ≤ 1
(sec θ)2 ≥ 1
(tan θ)2 < ∞
0 ≤ have
so want
so pick
1 − u2
0 ≤ 1 − u2
u2 ≤ 1
u = sin θ
u2 − 1
0 ≤ u2 − 1
1 ≤ u2
u = sec θ
u2 + 1
0 ≤ u2 + 1
−1 ≤ u2
u = tan θ
Recall:
if have
17.01.01 (yr.mn.dy)
then want
Page 1 of 2
Techniques of Integration
Trigonometric Substitution
Example: Evaluate the integral
The integrand
1
(4x2 +9)2
R
1
(4x2 +9)2
dx .
contains a u2 + a2 since
4x2 + 9 = (2x)2 + (3)2 = u2 + a2
where u = 2x and a = 3.
As suggested in the Summary Chart, we try the substitution u = a tan θ to get
2x = 3 tan θ
(sub)
2 dx = 3 sec2 θ dθ
(∗)
4x2 + 9 = (2x)2 + 32 = (3 tan θ)2 + 32 = 32 tan2 θ + 32 = 32 tan2 θ + 1 = 32 sec2 θ ,
where at (∗) we used 1 + tan2 θ = sec2 θ. Now we are ready to substitute into the original integral.
Z 3
Z
Z
Z
Z
sec2 θ dθ
3 1
sec2 θ dθ
1
dθ
1
dx
2
=
=
cos2 θ dθ
2 =
2 =
4
4θ
3
2θ
3
2
2
2
2
3
sec
2
·
3
sec
2
·
3
[4x + 9]
[3 sec θ]
.
now use a half ∠ formula: cos2 θ = 1+cos(2θ)
Z 2
Z
Z
1 1
1
=
(1 + cos(2θ)) dθ = 2 3
1 dθ + cos(2θ) dθ
2 · 33 2
2 ·3
Z
Z
1
1
= 2 3
1 dθ +
cos(2θ) (2dθ)
2 ·3
2
1
1
= 2 3 θ + sin(2θ) + C
2 ·3
2
We want our answer in terms of x, not θ. Using 2x = 3 tan θ, we can compute (a trig function)(θ) but
we have sin (2θ). To get rid of that unwanted 2, we use the double ∠ formula sin(2θ) = 2 sin θ cos θ.
1
1
= 2 3 θ + 2 sin θ cos θ + C
2 ·3
2
1
1
= 2 3 θ + 2 3 sin θ cos θ + C
2 ·3
2 ·3
How to find (the needed trig function)(θ)? Reference Triangle using original substitution.
2x = 3 tan θ
⇒
2x
opposite
opp.
adj.
=
. Know : sin θ =
& cos θ =
3
adjacent
hyp.
hyp.
p
√
32 + (2x)2 = 4x2 + 9
hyp.
2x
2x
θ
θ
3
3
⇒
tan θ =
1
3
2x
1
2x
= 2 3 arctan
+ 2 3 · √
· √
+C
2 ·3
3
2 ·3
4x2 + 9
4x2 + 9
1
2x
1
x
= 2 3 arctan
+
·
+C
2 ·3
3
2 · 32
4x2 + 9
1
2x
x
=
arctan
+
+C .
108
3
18(4x2 + 9)
17.01.01 (yr.mn.dy)
Page 2 of 2
Techniques of Integration