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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
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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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
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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