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need a hint, integrate sec^2(cosx) dx
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I think chain rule?
hm, how would I start it?
F'(x) = f'(g(x)) g'(x).
what is the derivative of sec^2(x) ?
2tan(x)sec^2(x)? but I'm trying to integrate
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Opps my bad..sorry
it's ok, any idea how to integrate this?
Change sec^2 to 1-tan^2 then convert equation to sin /cosine. sec^2(cosx) = 1 - tan^2(cosx) =[1-sin^2(cosx)/cos^2(cosx)]. . . . . . . . . . .
then Take t=cosx So dt=-sinxdx Thus, (1) becomes 1 - sin^2(t)/cos^2(t)-dt/sinx
alright, ty
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