Posted by Dartchen on August 29, 2002 at 14:22:25:
In Reply to: Some corrections.... posted by Subhotosh khan on August 29, 2002 at 12:53:05:
Doh! Hate it when I make stupid mistakes which make things simpler...guess there was a reason I felt something was wrong. Its been years since I've had a calculus class so I'll continue to pray to the math gods and demonstrate my failings on this problem.
Taking the corrected step of:
2 integral[u* (sin(u)/cos(u))]du )
It seems that my original idea for substitution is going to leave me with something I'm not sure how to solve for since I'd end up with:
-2 integral u*(dt/t)
Reading the faqs I don't see a way to integrate this as it stands and I can't think of a way to substitute for u (or x^1/2 for that matter) in relation to t.
Sorry...I know this is basic but I'm trying to relearn as fast as I can. If you know a reference or article on the net I can pull up that would give me direction I'll read it to try to keep from asking any more stupid questions. Thanks again for all the help!
: : Was given a problem that originally seemed simple but I think I'm making some conceptual errors on.
: : integral tan(x^1/2)dx
: : My original thoughts on the solution would be substitution
: : x^1/2 = u
: du = 1/2*x^(-1/2)dx = 1/(2u)dx
: dx = 2*u du
: : 2 integral[u* (sin(u)/cos(u))]du )
: Now of course rest of your "integration" need to be changed.
: : cos(u) = t
: : dt = -sin(u)du
: : sooo... -2integral dt/t = -2ln|t|+C
: : -2ln|cos(u)|+C
: : -2ln|cos(x^1/2)|+C ??
: : I'm probably way off but any hints/help would be GREATLY appreciated!!
: : Thanks!!
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