LECTURE 18: TRIGONOMETRIC SUBSTITUTIONS
MINGFENG ZHAO
February 23, 2015
Theorem 1.
Z
xp dx
Z
cos(ax) dx
Z
sin(ax) dx
Z
Z
Z
sec2 (ax) dx
csc2 (ax) dx
Z
eax dx
1
dx
− x2
Z
1
dx
2
x + a2
Z
tan(x) dx
Z
sec(x) dx
√
a2
=
 p+1

 x
+ C,
p+1

 ln |x| + C,
if p 6= −1
if p = −1.
1
sin(ax) + C
a
1
= − cos(ax) + C
a
1
=
tan(ax) + C
a
1
= − cot(ax) + C
a
1 ax
=
e +C
a
x
= sin−1
+C
a
x
1
=
tan−1
+C
a
a
=
=
− ln | cos(x)| + C
=
ln | sec(x) + tan(x)| + C
=
sec(x) + C.
Z
sec(x) tan(x) dx
Z
To evaluate
tanm (x) secn (x) dx:
Recall:
tan2 (x) + 1 = sec2 (x),
d
tan(x) = sec2 (x),
dx
1
and
d
sec(x) = sec(x) tan(x).
dx
2
MINGFENG ZHAO
Cases
Strategy
n even
Split off sec2 (x), rewrite the remaining even power of sec(x) in terms of tan(x),
and use u = tan(x)
m odd
Split off sec(x) tan(x), rewrite the remaining even power of tan(x) in terms of sec(x),
and use u = sec(x)
m even and n odd
Rewrite the even power of tan(x) in terms of sec(x) to produce a polynomial in sec(x),
apply reduction formula to each term
Z
Example 1. Evaluate
tan2 (x) sec(x) dx.
This is Case 3. Then
Z
Recall the reduction formula:
Z
tan2 (x) sec(x) dx
secn (x) dx =
Z
[sec2 (x) − 1] sec(x) dx
Z
Z
=
sec3 (x) dx − sec(x) dx
=
secn−2 (x) tan(x) n − 2
+
n−1
n−1
Z
secn−2 (x) dx.
Then we have
Z
2
tan (x) sec(x) dx
Z
=
sec(x) tan(x) 1
+
2
2
Z
sec(x) tan(x) 1
−
2
2
Z
=
=
sec(x) tan(x) 1
− ln | sec(x) + tan(x)| + C.
2
2
sec(x) dx −
sec(x) dx
sec(x) dx
Integrals involving a2 − x2
Let x = a sin(θ), then
a2 − x2 = a2 − a2 sin2 (θ) = a2 cos2 (θ).
Z
dx
3 .
(16 − x2 )2 x
Let x = 4 sin(θ), then θ = sin−1
and dx = 4 cos(θ)dθ. So we have
4
Z
Z
dx
1
=
3
3 · 4 cos(θ) dθ
(16 − x2 ) 2
(16 − 16 sin2 (θ)) 2
Example 2. Evaluate
Let x = 4 sin(θ)
LECTURE 18: TRIGONOMETRIC SUBSTITUTIONS
Z
4 cos(θ)
=
3
[16 cos2 (θ)] 2
dθ
Z
4 cos(θ)
dθ
43 cos3 (θ)
Z
1
1
dθ
16
cos2 (θ)
Z
1
sec2 (θ) dθ
16
1
tan(θ) + C.
16
=
=
=
=
Since sin(θ) =
x
, then
4
tan(θ) = √
x
.
16 − x2
Then we get
Z
dx
(16 −
3
x2 ) 2
=
x
√
+ C.
16 16 − x2
Integrals involving a2 + x2
Let x = a tan(θ), then
a2 + x2 = a2 + a2 tan2 (θ) = a2 sec2 (θ).
Z
dx
.
x2 + 4
Let x = 2 tan(θ), then dx = 2 sec2 (θ)dθ. So we have
Z
Z
dx
2 sec2 (θ)
√
p
=
dθ
x2 + 4
4 tan2 (θ) + 4
Z
2 sec2 (θ)
dθ
=
2 sec(θ)
Z
=
sec(θ) dθ
Example 3. Evaluate
√
=
Since tan(θ) =
Let x = 2 tan(θ)
ln | sec(θ) + tan(θ)| + C.
x
, then
2
√
sec(θ) =
4 + x2
.
2
So we get
Z
dx
√
x2 + 4
=
√
4 + x2
x ln + +C
2
2
3
4
MINGFENG ZHAO
p
ln x2 + 4 + x − ln 2 + C
p
= ln x2 + 4 + x + C.
=
Integrals involving x2 − a2
Let x = a sec(θ), then
x2 − a2 = a2 sec2 (θ) − a2 = a2 tan2 (θ).
4
√
x2 + 4x − 5
dx.
x+2
1
Notice that x2 + 4x − 5 = (x + 2)2 − 22 − 5 = (x + 2)2 − 9. Then
Z 4p
Z 4√ 2
(x + 2)2 − 9
x + 4x − 5
dx =
dx
x+2
x+2
1
1
Z 6√ 2
u −9
=
du Let u = x + 2.
u
3
Z √ 2
u −9
For
du, let u = 3 sec(θ), then du = 3 sec(θ) tan(θ)dθ. So we get
u
Z p
Z √ 2
9 sec2 (θ) − 9
u −9
du =
· 3 sec(θ) tan(θ) dθ Let u = 3 sec(θ)
u
3 sec(θ)
Z
3 tan(θ)
=
· 3 sec(θ) tan(θ) dθ
3 sec(θ)
Z
= 3 tan2 (θ) dθ
Z
= 3 [sec2 (θ) − 1] dθ
Z
Example 4. Evaluate
=
3[tan(θ) − θ] + C.
3
u
Since sec(θ) = , then cos(θ) = , θ = cos−1
3
u
3
and
u
√
tan(θ) =
u2 − 9
.
3
Then we get
Z
1
4
√
x2 + 4x − 5
dx
x+2
6
Z
√
=
3
"√
=
3
u2 − 9
du
u
u
u2 − 9
− sec−1
3
3
#6
3
LECTURE 18: TRIGONOMETRIC SUBSTITUTIONS
5
"√
#
27
1
= 3
− 3 0 − cos−1 (1)
− cos−1
3
2
√
π
= 3
−0
3−
3
√
= 3 3 − π.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca