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(c) Evaluate: sec" (x)dz.

(c) Evaluate: sec" (x)dz.

Advanced Engineering Mathematics
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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Question
Do ONE of the following: (a) Bacterial population A is growing exponentially while B is shrinking exponentially. At time t-0, population A has 3000 cells and population B has 9000 cells. If the two populations have the same number of cells at t=5 what is the ratio of population A to population B at t=15? (b) Find the volume bound by the curves x = 2y? – 3y – 3 and a = = -y? +3 if they are rotated around the line y = -2. (c) Evaluate: /sec" (a)dz .
Transcribed Image Text:Do ONE of the following: (a) Bacterial population A is growing exponentially while B is shrinking exponentially. At time t-0, population A has 3000 cells and population B has 9000 cells. If the two populations have the same number of cells at t=5 what is the ratio of population A to population B at t=15? (b) Find the volume bound by the curves x = 2y? – 3y – 3 and a = = -y? +3 if they are rotated around the line y = -2. (c) Evaluate: /sec" (a)dz .
Do ONE of the following: (a) Bacterial population A is growing exponentially while B is shrinking exponentially. At time t-0, population A has 3000 cells and population B has 9000 cells. If the two populations have the same number of cells at t=5 what is the ratio of population A to population B at t=15? (b) Find the volume bound by the curves x = 2y? – 3y – 3 and a = = -y? +3 if they are rotated around the line y = -2. (c) Evaluate: /sec" (a)dz .
Transcribed Image Text:Do ONE of the following: (a) Bacterial population A is growing exponentially while B is shrinking exponentially. At time t-0, population A has 3000 cells and population B has 9000 cells. If the two populations have the same number of cells at t=5 what is the ratio of population A to population B at t=15? (b) Find the volume bound by the curves x = 2y? – 3y – 3 and a = = -y? +3 if they are rotated around the line y = -2. (c) Evaluate: /sec" (a)dz .
Expert Solution
Step 1

(c)

reduction formula for sec x  is given as,

In(n-1) =secn-2x tanx +(n-2)In-2In = ∫secnx dx    where 

taking n = 5

I5×4 = sec3x tanx +3I3

Also I3 can be resolved in the same way as

I3 ×2=secx tanx +I1

Plugging this value above

4I5 = sec3x tanx +312secx tanx+12I1

Further,

I5 =14 sec3x tanx +38secx tanx+∫sec x dx

 

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