Logarithmic Functions Continued
Joseph Lee
Metropolitan Community College
Joseph Lee
Logarithmic Functions Continued
Antiderivative of
1
x
Previously, we learned
d
1
[ln |x|] = . Therefore,
dx
x
Z
1
dx = ln |x| + C .
x
Joseph Lee
Logarithmic Functions Continued
Example 1.
Z
Evaluate the indefinite integral
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1
dx.
2x − 3
Logarithmic Functions Continued
Example 1.
Z
Evaluate the indefinite integral
1
dx.
2x − 3
Solution. Let u = 2x − 3. Then du = 2dx.
Z
Z
1
1
2
dx =
dx
2x − 3
2 2x − 3
Joseph Lee
Logarithmic Functions Continued
Example 1.
Z
Evaluate the indefinite integral
1
dx.
2x − 3
Solution. Let u = 2x − 3. Then du
Z
Z
1
1
dx =
2x − 3
2
Z
1
=
2
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= 2dx.
2
dx
2x − 3
du
u
Logarithmic Functions Continued
Example 1.
Z
Evaluate the indefinite integral
1
dx.
2x − 3
Solution. Let u = 2x − 3. Then du
Z
Z
1
1
dx =
2x − 3
2
Z
1
=
2
=
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= 2dx.
2
dx
2x − 3
du
u
1
ln |u| + C
2
Logarithmic Functions Continued
Example 1.
Z
Evaluate the indefinite integral
1
dx.
2x − 3
Solution. Let u = 2x − 3. Then du
Z
Z
1
1
dx =
2x − 3
2
Z
1
=
2
= 2dx.
2
dx
2x − 3
du
u
=
1
ln |u| + C
2
=
1
ln |2x − 3| + C
2
Joseph Lee
Logarithmic Functions Continued
Antiderivatives of sin x and cos x
Z
sin x dx = − cos x + C .
Z
cos x dx = − sin x + C .
Joseph Lee
Logarithmic Functions Continued
Example 2.
Z
Evaluate the indefinite integral
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tan x dx.
Logarithmic Functions Continued
Example 2.
Z
Evaluate the indefinite integral
Solution.
tan x dx.
Z
Z
tan x dx =
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sin x
dx
cos x
Logarithmic Functions Continued
Example 2.
Z
Evaluate the indefinite integral
Solution.
tan x dx.
Z
Z
tan x dx =
sin x
dx
cos x
Let u = cos x. Then du = − sin x dx.
Joseph Lee
Logarithmic Functions Continued
Example 2.
Z
Evaluate the indefinite integral
Solution.
tan x dx.
Z
Z
tan x dx =
sin x
dx
cos x
Let u = cos x. Then du = − sin x dx.
Z
Z
du
sin x
dx = −
cos x
u
Joseph Lee
Logarithmic Functions Continued
Example 2.
Z
Evaluate the indefinite integral
Solution.
tan x dx.
Z
Z
tan x dx =
sin x
dx
cos x
Let u = cos x. Then du = − sin x dx.
Z
Z
du
sin x
dx = −
cos x
u
= − ln |u| + C
Joseph Lee
Logarithmic Functions Continued
Example 2.
Z
Evaluate the indefinite integral
Solution.
tan x dx.
Z
Z
tan x dx =
sin x
dx
cos x
Let u = cos x. Then du = − sin x dx.
Z
Z
du
sin x
dx = −
cos x
u
= − ln |u| + C
= − ln | cos x| + C
Joseph Lee
Logarithmic Functions Continued
Example 3.
Z
Evaluate the indefinite integral
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cot x dx.
Logarithmic Functions Continued
Example 3.
Z
Evaluate the indefinite integral
Solution.
cot x dx.
Z
Z
cot x dx =
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cos x
dx
sin x
Logarithmic Functions Continued
Example 3.
Z
Evaluate the indefinite integral
Solution.
cot x dx.
Z
Z
cot x dx =
cos x
dx
sin x
Let u = sin x. Then du = cos x dx.
Joseph Lee
Logarithmic Functions Continued
Example 3.
Z
Evaluate the indefinite integral
Solution.
cot x dx.
Z
Z
cot x dx =
cos x
dx
sin x
Let u = sin x. Then du = cos x dx.
Z
Z
cos x
du
dx =
sin x
u
Joseph Lee
Logarithmic Functions Continued
Example 3.
Z
Evaluate the indefinite integral
Solution.
cot x dx.
Z
Z
cot x dx =
cos x
dx
sin x
Let u = sin x. Then du = cos x dx.
Z
Z
cos x
du
dx =
sin x
u
= ln |u| + C
Joseph Lee
Logarithmic Functions Continued
Example 3.
Z
Evaluate the indefinite integral
Solution.
cot x dx.
Z
Z
cot x dx =
cos x
dx
sin x
Let u = sin x. Then du = cos x dx.
Z
Z
cos x
du
dx =
sin x
u
= ln |u| + C
= ln | sin x| + C
Joseph Lee
Logarithmic Functions Continued
Example 4.
Z
Evaluate the indefinite integral
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sec x dx.
Logarithmic Functions Continued
Example 4.
Z
Evaluate the indefinite integral
sec x dx.
Solution.
Z
Z
sec x dx =
sec x
sec x + tan x
sec x + tan x
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Z
dx =
sec2 x + sec x tan x
dx
sec x + tan x
Logarithmic Functions Continued
Example 4.
Z
Evaluate the indefinite integral
sec x dx.
Solution.
Z
Z
sec x dx =
sec x
sec x + tan x
sec x + tan x
Z
dx =
sec2 x + sec x tan x
dx
sec x + tan x
Let u = sec x + tan x. Then du = (sec x tan x + sec2 x) dx.
Joseph Lee
Logarithmic Functions Continued
Example 4.
Z
Evaluate the indefinite integral
sec x dx.
Solution.
Z
Z
sec x dx =
sec x
sec x + tan x
sec x + tan x
Z
dx =
sec2 x + sec x tan x
dx
sec x + tan x
Let u = sec x + tan x. Then du = (sec x tan x + sec2 x) dx.
Z
sec2 x + sec x tan x
dx =
sec x + tan x
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Z
du
u
Logarithmic Functions Continued
Example 4.
Z
Evaluate the indefinite integral
sec x dx.
Solution.
Z
Z
sec x dx =
sec x
sec x + tan x
sec x + tan x
Z
dx =
sec2 x + sec x tan x
dx
sec x + tan x
Let u = sec x + tan x. Then du = (sec x tan x + sec2 x) dx.
Z
sec2 x + sec x tan x
dx =
sec x + tan x
Z
du
u
= ln |u| + C
Joseph Lee
Logarithmic Functions Continued
Example 4.
Z
Evaluate the indefinite integral
sec x dx.
Solution.
Z
Z
sec x dx =
sec x
sec x + tan x
sec x + tan x
Z
dx =
sec2 x + sec x tan x
dx
sec x + tan x
Let u = sec x + tan x. Then du = (sec x tan x + sec2 x) dx.
Z
sec2 x + sec x tan x
dx =
sec x + tan x
Z
du
u
= ln |u| + C
= ln | sec x + tan x| + C
Joseph Lee
Logarithmic Functions Continued
Example 5.
Z
Evaluate the indefinite integral
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csc x dx.
Logarithmic Functions Continued
Example 5.
Z
Evaluate the indefinite integral
csc x dx.
Solution.
Z
Z
csc x dx =
csc x
csc x + cot x
csc x + cot x
Joseph Lee
Z
dx =
csc2 x + csc x cot x
dx
csc x + cot x
Logarithmic Functions Continued
Example 5.
Z
Evaluate the indefinite integral
csc x dx.
Solution.
Z
Z
csc x dx =
csc x
csc x + cot x
csc x + cot x
Z
dx =
csc2 x + csc x cot x
dx
csc x + cot x
Let u = csc x + cot x. Then du = −(csc x cot x + cot2 x) dx.
Joseph Lee
Logarithmic Functions Continued
Example 5.
Z
Evaluate the indefinite integral
csc x dx.
Solution.
Z
Z
csc x dx =
csc x
csc x + cot x
csc x + cot x
Z
dx =
csc2 x + csc x cot x
dx
csc x + cot x
Let u = csc x + cot x. Then du = −(csc x cot x + cot2 x) dx.
Z
Z
csc2 x + csc x cot x
du
dx = −
csc x + cot x
u
Joseph Lee
Logarithmic Functions Continued
Example 5.
Z
Evaluate the indefinite integral
csc x dx.
Solution.
Z
Z
csc x dx =
csc x
csc x + cot x
csc x + cot x
Z
dx =
csc2 x + csc x cot x
dx
csc x + cot x
Let u = csc x + cot x. Then du = −(csc x cot x + cot2 x) dx.
Z
Z
csc2 x + csc x cot x
du
dx = −
csc x + cot x
u
= − ln |u| + C
Joseph Lee
Logarithmic Functions Continued
Example 5.
Z
Evaluate the indefinite integral
csc x dx.
Solution.
Z
Z
csc x dx =
csc x
csc x + cot x
csc x + cot x
Z
dx =
csc2 x + csc x cot x
dx
csc x + cot x
Let u = csc x + cot x. Then du = −(csc x cot x + cot2 x) dx.
Z
Z
csc2 x + csc x cot x
du
dx = −
csc x + cot x
u
= − ln |u| + C
= − ln | csc x + cot x| + C
Joseph Lee
Logarithmic Functions Continued
Textbook Exercises
Exercise 6
Exercise 12
Exercise 18
Exercise 24
Exercise 28
Exercise 36
Exercise 52
Exercise 72
Joseph Lee
Logarithmic Functions Continued