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Answer Key for Quiz 2 (section A)
- We can let u = ln(ln x) and dv = dxx , in which case v = ln x (since x is positive) and du =
ln x
x
dx.
Then ∫ ln (ln x)^1 x
dx = ln(ln x) ln x −
ln x dx x ln x
= ln(ln x) ln x −
dx x = ln(ln x) ln x − ln x + C = (ln x) [ln(lnx) − 1] + C.
Or we can start by substituting w = ln x, in which case dw = (^1) x dx and we have
∫ ln (ln x)
x dx^ =
ln w dw.
This one we’ve done before: integrate by parts with u = ln w, dv = dw, so that v = w and du = dww. Then
∫ ln (ln x)^1 x
dx =
ln w dw
= w ln w −
w
dw w =^ w^ ln^ w^ −
dw
= w ln w − w + C = ln x ln(ln x) − ln x + C.
- In both of (^) ∫ arcsec x dx and
x arcsec x dx
we want to integrate by parts with u = arcsec x, so that du can be 1 x
x^2 − 1
dx. In the first case dv = dx,
so v = x and we have ∫ arcsec x dx = x arcsec x −
x dx x
x^2 − 1
= x arcsec x −
√dx x^2 − 1
In the second case dv = x dx, so that v = x
2 2 and we have ∫ x arcsec x dx =
x^2 2 arcsec^ x^ −
x^2 2
dx x
x^2 − 1
x^2 2 arcsec^ x^ −^
√x dx x^2 − 1
This is easier than the first case, because the substitution w = x^2 − 1 will work here. We have dw = 2x dx, so dw 2 = x dx and we get
∫ x arcsec x dx =
x^2 2 arcsec^ x^ −^
√x dx x^2 − 1
=
x^2 2
arcsec x −
∫ (^) dw √^2 w
=
x^2 2
arcsec x −
w−^
(^12) dw
= x
2 2
arcsec x − 1 4
2 w 12 + C
=
x^2 2
arcsec x −
x^2 − 1 + C.
The substitution y =
x^2 − 1 is even a little better. Then
dy =
(x^2 − 1)−^
(^12) 2 x dx = √x dx x^2 − 1
so ∫ x arcsec x dx =
x^2 2
arcsec x −
x dx √ x^2 − 1
=
x^2 2
arcsec x −
dy =
x^2 2
arcsec x −
y 2
+ C
x^2 2 arcsec^ x^ −^
x^2 − 1 + C.
The same substitution works in the first case, but in a very tricky way. We saw above that
if y =
x^2 − 1 , then dy = √x dx x^2 − 1
x dx y
, or
dy x
dx y
A property of fractions which is not so well known, although easy to prove, is that
if a b
= c d
, then a b
= c d
= a^ +^ c b + d
Therefore (^) ∫ √dx x^2 − 1
dx y =
dx + dy x + y
with y =
x^2 − 1 as above. If we now let z = x + y, then dz = dx + dy and we get ∫ √dx x^2 − 1
dx + dy x + y
dz z = ln |z| + C = ln |x + y| + C = ln
∣x^ +^
x^2 − 1
∣ +^ C.
This is a difficult integral, unless you happen to know that it equals a function we haven’t discussed, the inverse hyperbolic cosine of x. Another way of doing it is to let x = sec θ, in which case dx = sec θ tan θ dθ and we have ∫ dx √ x^2 − 1
sec θ tan θ dθ √ sec^2 θ − 1
sec θ tan θ dθ tan θ
=
sec θ dθ = ln | sec θ + tan θ| + C
= ln
∣x +
x^2 − 1
∣ + C.
If we can do this integral, then we can say
∫ arcsec x dx = x arcsec x −
√dx x^2 − 1 = x arcsec x − ln
x +
x^2 − 1
+ C,
where we can drop the absolute values since we assumed at the start that x is positive.
