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[apcalculus] Re: Integral Problem  Finney DeMana #59, p. 575
Posted:
Apr 5, 2005 9:52 AM


Hello George,
The answers are the same, since they differ by a constant. Remember
(1/2)tan^2 u = (1/2)sec^2 u  1/2.
Hope this helps.
Dick Maher
Richard J. Maher Mathematics and Statistics Loyola University Chicago 6525 N. Sheridan Rd. Chicago, Illinois 60626 17735083565 rjm@math.luc.edu
On Mon, 4 Apr 2005, George Becker wrote:
This problem asks for the integral of sin(x3)/cos^3(x3)dx. Using substitution with u=cosx yields 1/(2cos^2(x3))+ C which is the answer the book provides. Alternatively, this could be written as 1/2(sec^2(x3)).
Question is when you change the original integral to tan(x3)sec^2(x3)dx and then integrate by substitution with u=tanx, you get 1/2(tan^2(x3))+ C which does not appear to be the same answer, even using common trig identities.
Why does these two answers not agree?
George Becker Pope John HS Sparta, NJ
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