Extra Problems
Name:
1. Prove the following reduction formula:
Z
Z
secn−2 (x) tan(x) n − 2
n
+
secn−2 (x) dx
sec (x) dx =
n−1
n−1
R
Proof. This requires two application of by-parts. Let In = secn (x) dx, we now using
integration by part for In to find a relation with In−2 . First recall that:
0
sec x = sec x tan x
sec2 x = tan2 x + 1

n−2

u = sec (x)
Then
du = (n − 2) secn−3 (x) sec(x) tan(x) dx


= (n − 2) secn−2 (x) tan(x)
(
dv = sec2 (x) dx
v = tan(x)
Using integrating by part:
Z
tan x (n − 2) secn−2 x tan x dx
Z
n−2
2
= sec
x tan x − (n − 2) secn−2 x tan
| {z x} dx
n−2
In = sec
x tan x −
sec2 x−1
"Z
= secn−2 x tan x − (n − 2)
"Z
= secn−2 x tan x − (n − 2)
n−2
2
sec
x sec x − 1 dx
In = sec
#
Z
secn x dx − secn−2 x dx
|
{z
} |
{z
}
In
n−2
#
secn x − secn−2 x dx
"Z
= secn−2 x tan x − (n − 2)
#
In−2
x tan x − (n − 2)In + (n − 2)In−2
Thus we must have:
In + (n − 2)In = secn−2 x tan x + (n − 2)In−2
Thus
(n − 1)In = secn−2 x tan x + (n − 2)In−2
Devide two sides by (n − 1) we obtain the final formula
Z
Z
secn−2 (x) tan(x) n − 2
n
sec (x) dx =
+
secn−2 (x) dx
n−1
n−1
Extra Problems
Name:
Remark. The alternative method for substitution:
u = sec x
1
u = U (t) =
2
1
t+
t
u = tan x
1
u = V (t) =
2
1
t−
t
and
Some useful inequalities for U and V :
U 2 (t) = V 2 (t) + 1
1
U 0 (t)
=
V (t)
t
Z
2. Using either way (x = sec θ – x = U (t)to find:
t = U (t) + V (t)
√
1
x2 − 1
dx.
Proof 1. Let x = sec θ then dx = sec θ tan θ dθ, so
Z
Z
Z
Z
sec θ tan θ
sec θ tan θ
1
√
√
dx =
dθ =
dθ = sec θ dθ
tan θ
x2 − 1
sec2 −1
= ln |sec x + tan x| + C
For the strategy to find
Canvas page.
R
sec x dx, see ”Tricky” way to integrate secant(x) in your
Proof 2. Let x = U (t), then dx = U 0 (t) dt, so:
Z
Z
Z 0
Z
U 0 (t)
1
U (t)
1
√
p
dx =
dt =
dt =
dt = ln(|t|) + C
V (t)
t
x2 − 1
U 2 (t) − 1
Now notice that x = U (t) then since U 2 (t) − 1 = V (t), we have V (t) =
√
t = U (t) + V (t) = x + x2 − 1
and thus:
Z
√
1
x2 − 1
dx = ln(|t|) + C = ln(|x +
√
x2 − 1|) + C
√
x2 − 1, thus: