Use an appropriate substitution and then a trigonometric substitution to evaluate the integral. In 5 ļ 0 e¹dt √e²t +36 Which substitution transforms the given integral into one that can be evaluated directly in terms of 0? O A. et = 6 sec 0 OB. e-6 sin 0 C. e¹=6tan 0 Find the correct expression for the differential. e¹ dt = sec ²0 de
Use an appropriate substitution and then a trigonometric substitution to evaluate the integral. In 5 ļ 0 e¹dt √e²t +36 Which substitution transforms the given integral into one that can be evaluated directly in terms of 0? O A. et = 6 sec 0 OB. e-6 sin 0 C. e¹=6tan 0 Find the correct expression for the differential. e¹ dt = sec ²0 de
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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Transcribed Image Text:Use an appropriate substitution and then a trigonometric substitution to evaluate the integral.
In 5
S
0
et dt
e2t
+ 36
Which substitution transforms the given integral into one that can be evaluated directly in terms of 0?
A. et = 6 sec 0
B. et = 6 sin 0
et = 6 tan 0
Find the correct expression for the differential.
2
et dt =
t=sec ²0 de
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