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Old 06-15-2007   #2
YvesBenoit
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Ok I worked at the first problem and figured it out.

using the identity 1/sinx = cscx...

(2/(sinx)^2)x = 2(csc^2)x

Using the identity cosx/sinx = cotx

3cosx/((sinx)^2) = 3cotx/sinx

And using the identity 1/sinx = cscx

3cotx/sinx = 3cscxcotx

Giving a final developed equation of
(integral) (2(csc^2)x + 3cscxcotx + sinx) dx

Remembering these derivatives
cosx = -sinx
cscx = -cscxcotx
tanx = (csc^2)x

Final integrated answer = -2cotx - 3cscx - cosx + C

C being any integer.

But I still can't figure out the second one, although it's probably much easier