Calculus II
Practice Problems 3: Answers
1. Differentiate:
f x 2x 6
3x 5
Answer. This problem is here to suggest a different way, called logarithmic differentiation, of differentiating
expressions like this. First we rewrite the function exponentially:
1
f x 2x 6
3x 5
1 2
1 2
and then we take the logarithm:
ln f x Now differentiate this equation:
f x
f x
1
ln 2x 6 ln 3x 5 2
1
2
2 2x 6
3
3x 5
and now multiply through by f x , as in equation (1):
f x 2. a)
b)
f x
1
2
2 2x 6
3
3x 5
2x 6
3
2x 6 1 2 3x 5 3 2
1 2 3x 5 1 2 2
tan arccos x 1
x2
tan2 arccosx Answer. a. Draw a right triangle with hypotenuse 1, an angle α , and label the side adjacent to α as x so that
α arccos x (see the figure).
1
PSfrag replacements
α
x
Now, by the Pythagorean theorem, the side opposite α has length 1 x 2 . Thus
tan arccosx tan α
1 x2
x
1 x2
Now, for b):
1
x2
1
x2
tan arccos x 2
1 x2
x
2
1
1 x2 x2
1
From the diagram, 1 x sec α , so this also follows from the identity tan 2 α 1 sec2 α .
3. Differentiate y arccos x
Answer. By the chain rule:
y
4. Differentiate:
g x arcsin ln x 1
1
x
1
2 x
2
1
2 x 1 x
Answer. By the chain rule:
g x x
1
1
ln x 2
5. Integrate
2
xdx
1 4x2
2
dx
2
0 1 4x
a)
b)
0
Answer. For a), let u 1 4x 2 du 8xdx. Then
2
0
xdx
1 4x2
1
8
17
1
du
u
1
17
lnu 1
8
ln17
8
For b), however, we should make the substitution: u 2x du 2dx. Then
2
0
dx
1 4x2
1
2
4
0
du
1 u2
1
4
arctanu 0
2
arctan4
2
This problem illustrates that one must be careful in deciding what substitution to make. One has to scan the
integrand to see how to write it as a product of a function of u (the substitution to be made) and the differential
of u.
6. Integrate:
a)
b)
ex dx
1
dx
x
e e
x
e2x
Answer. For part a) recognize that for u e x du ex dx, and thus
ex dx
e2x 1
du
1 u2
arctanu C arctanex C
For part b), we see that, in order to make the same substitution, we need an e x in the numerator. So put it
there, by multiplying by e x ex 1:
dx
x
e e
x
ex dx
ex ex e x
ex dx
e2x 1
arctanex C by part a).
7.
dx
5 4x x2
Answer. To get this to look like something recognizable, we complete the square: 5 4x x 2
The problem now looks like
dx
9
x 2
2
9 x 2 2.
This looks like the integral should involve arccos, but we need a 1 where the 9 is. We fix that by the substitution 3u x 2 3du dx. Then we have
x 2
3du
du
arccos u C arccos
C
2
2
3
9 9u
1 u
9 x 2
dx
2
8. Integrate:
a)
tan2 xdx Answer. tan2 x sec2 x 1, so
tan2 xdx b)
sec2 x 1 dx tanx x C
tan3 xdx Answer. Using the same identity
tan3 xdx sec2 x 1 tanxdx sec x sec x tan x dx tan xdx
sec2 x
ln cos x C 2
sec x tan xdx for the first integral.
using the substitution u sec x du 9.
x tan x2 1 dx Answer. Let u x2 1 du 2xdx. Then
x tan x2 1 dx 1
1
1
tan udu ln cos u C ln cos x2 1 C
2
2
2
10
dx
x2 6x 13
By completing the square, we see that x 2 6x 13 x2
dx
6x 13
x 3
x
2
4. Let u x 3 du dx, so that we have
dx
3 2 4
du
u2
4
Now, we know the integral of du u 2 1 ; but how do we do du u 2 4 ? Again a little algebra comes into
play:
u 2
u2 4 4
1
2
Try the substitution v u 2 dv du 2:
du
u2 4
1
4
2dv
v2 1
1
arctanv C 2
1
x 3
arctan
2
2
C