Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:Evaluate the integral.
S
arctan(3t) dt
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![Step 1
To use the integration-by-parts formula
Step 2
Choose u = arctan(3t). This means that dv =
Now, since u = arctan(3t), then du =
Step 3
With our choice that dv= dt, then v =
Judv.
fu
dv = uv - v du
= t arctan(3t)
u dv = uv -
3
1+ (3t)
2
Step 4
Now, the integration-by-parts formula gives us
[1 dt
1 dt = 0
1₁
3t
1 + 9t²
-]₁
dt.
v du, we must choose one part of
F/
arctan(3t) dt to be u, with the rest becoming dv.
Choosing w = 1 + 9t2, we differentiate to find dw = 18
3
9t² + 1
dt
dt.
t
18 t dt.](https://content.bartleby.com/qna-images/question/37c407f4-5be5-4e70-b81d-676b039d3ff5/e1af2bd7-9943-4497-9acb-53f13c0ff066/24s24y_processed.png)
Transcribed Image Text:Step 1
To use the integration-by-parts formula
Step 2
Choose u = arctan(3t). This means that dv =
Now, since u = arctan(3t), then du =
Step 3
With our choice that dv= dt, then v =
Judv.
fu
dv = uv - v du
= t arctan(3t)
u dv = uv -
3
1+ (3t)
2
Step 4
Now, the integration-by-parts formula gives us
[1 dt
1 dt = 0
1₁
3t
1 + 9t²
-]₁
dt.
v du, we must choose one part of
F/
arctan(3t) dt to be u, with the rest becoming dv.
Choosing w = 1 + 9t2, we differentiate to find dw = 18
3
9t² + 1
dt
dt.
t
18 t dt.
Transcribed Image Text:Step 5
The integral
S
W
6
3t
1 + 9t²
1
By substitution we have
3
18
1 + 9t²
dt can be rewritten as
In(1 +
dw
=
=
t 3
1
6
18
Writing this in terms of t we have
J
dt.
6
t²).
W
dw
In(w).
Solution
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