Question

proove by integraion that (1/2)arctan(2x/1-2X) and arctan(1/x) are both integrals of the same function

Answer 1

think u mean by differentiaion

Answer 2

ok help me out please

Answer 3

no it cant be by differentiation as we r tryin to roll back the two integrals but ignore the c in F(x) + \[c \int\limits_{?}^{?}\]

Answer 4

F(x) + c

Answer 5

so the first is \[\frac{1}{2}\tan^{-1} (\frac{2x}{1-2x})\]

Answer 6

dont think its possbile

they must only differ by a constant

Answer 7

[S] [tan^-1 (2x/1-2X)]/2 = [S] tan^-1(1/x) right?

Answer 8

prob try some sum of angle formulas, possibly, but I dnt think it will make a difference

Answer 9

thanx guys.....yes they have to differ by a constant.....its time consuming whew......how can i master induction of inequalities?

Answer 10

huh, maybe diff of angles formula

Answer 11

tan^-1(1/x) = u

-1/x^2
--------- dx = du perhaps
1 + 1/x^2

Answer 12

tan(u) = 1/x is what i was thinking lol; lost track mid type

Answer 13

the difference is an angle of \[\pi/2\]

Answer 14

Are you sure you've stated the problem correctly? If you meant \[1/2\arctan(2x/1-x^2)\]
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)

Answer 15

eash man my bad ey guess you are a geenie

Answer 16

what, \[\frac{1}{2}\tan^{-1} (\frac{2x}{1-x^2}) \neq \tan^{-1} (\frac{1}{x})\]

Answer 17

You 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/(1-x^2) ==
(x*ArcTan[(2*x)/(1 - x^2)] - Log[1 + x^2])/2

Integrate 1/2tan^-1(2x/(1-2x) ==
(ArcTan[1 - 4*x]/4 + x*ArcTan[(2*x)/(1 - 2*x)] - Log[1 - 4*x + 8*x^2]/8)/2

Answer 18

^ 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