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anonymous
 5 years ago
proove by integraion that (1/2)arctan(2x/12X) and arctan(1/x) are both integrals of the same function
anonymous
 5 years ago
proove by integraion that (1/2)arctan(2x/12X) and arctan(1/x) are both integrals of the same function

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0think u mean by differentiaion

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok help me out please

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it cant be by differentiation as we r tryin to roll back the two integrals but ignore the c in F(x) + \[c \int\limits_{?}^{?}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the first is \[\frac{1}{2}\tan^{1} (\frac{2x}{12x})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont think its possbile they must only differ by a constant

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1[S] [tan^1 (2x/12X)]/2 = [S] tan^1(1/x) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0prob try some sum of angle formulas, possibly, but I dnt think it will make a difference

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanx guys.....yes they have to differ by a constant.....its time consuming whew......how can i master induction of inequalities?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0huh, maybe diff of angles formula

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1tan^1(1/x) = u 1/x^2  dx = du perhaps 1 + 1/x^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1tan(u) = 1/x is what i was thinking lol; lost track mid type

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the difference is an angle of \[\pi/2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you sure you've stated the problem correctly? If you meant \[1/2\arctan(2x/1x^2)\] ie x^2 rather than 2x on the bottom line, then the two expressions are identical, because tan2x = 2tan(x/2)/(1+tan^2(x/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0eash man my bad ey guess you are a geenie

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what, \[\frac{1}{2}\tan^{1} (\frac{2x}{1x^2}) \neq \tan^{1} (\frac{1}{x})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are quite right. Again! I'm not there yet, but using brute force with Mathematica might give us some ideas. I checked 2x on the bottom line as well as x^2: Integrate[ArcTan[1/x], x] == x*ArcCot[x] + Log[1 + x^2]/2 Integrate 1/2tan^1(2x/(1x^2) == (x*ArcTan[(2*x)/(1  x^2)]  Log[1 + x^2])/2 Integrate 1/2tan^1(2x/(12x) == (ArcTan[1  4*x]/4 + x*ArcTan[(2*x)/(1  2*x)]  Log[1  4*x + 8*x^2]/8)/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0^ I wouldnt worry about thinking about it, it was a bad question to begin with, I am very sure there was a problem with it
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