Find L 1- A = cos(x) (p) 2 Hint. There may be a limit of the bound to evaluate at infinity, do that at the very last step (Recall that is usually how we evaluate improper integrals, integrate up to some b first, then take the limit as b→ ∞o. Also, try not to split the limit up. Sometimes when you split a limit you would get a sum of two divergent term, while the limit of the sum may actually converge. So "combine them", then take the limit.) If you do it correctly, you should be able to write it in the form A ln(1 + p) where and N= (enter all in most reduced fractions)
Find L 1- A = cos(x) (p) 2 Hint. There may be a limit of the bound to evaluate at infinity, do that at the very last step (Recall that is usually how we evaluate improper integrals, integrate up to some b first, then take the limit as b→ ∞o. Also, try not to split the limit up. Sometimes when you split a limit you would get a sum of two divergent term, while the limit of the sum may actually converge. So "combine them", then take the limit.) If you do it correctly, you should be able to write it in the form A ln(1 + p) where and N= (enter all in most reduced fractions)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Hint. There may be a limit of the bound to evaluate at infinity, do that at the very last step (Recall
that is usually how we evaluate improper integrals, integrate up to some b first, then take the limit
as b→ ∞o. Also, try not to split the limit up. Sometimes when you split a limit you would get a
sum of two divergent term, while the limit of the sum may actually converge. So "combine them",
then take the limit.)
If you do it correctly, you should be able to write it in the form A ln(1 + p) where
A =
and N=
(enter all in most reduced fractions)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd8d776ae-c221-44f7-bbf2-f74f003dec75%2F9901d6fb-d5d6-486a-aa53-25facf02dcca%2Febb33rt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1-cos(x)
2
Find L
-](p)
Hint. There may be a limit of the bound to evaluate at infinity, do that at the very last step (Recall
that is usually how we evaluate improper integrals, integrate up to some b first, then take the limit
as b→ ∞o. Also, try not to split the limit up. Sometimes when you split a limit you would get a
sum of two divergent term, while the limit of the sum may actually converge. So "combine them",
then take the limit.)
If you do it correctly, you should be able to write it in the form A ln(1 + p) where
A =
and N=
(enter all in most reduced fractions)
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