insane maths problem - help (1 Viewer)
- Thread starter fatmike93
- Start date
i have a feeling that its wrong. the line where you sub 'u' in.
you changed it from 1+tanx (sqaured) / tanx to 1+ u (squared) / u.
so you basically let u = tanx as you subbed it in. however, you initially stated that u=ln(tanx). so...
maybe i'm wrong, but just check it. i dont think its that easy otherwise i woulda got it ages ago. LOL i asked my teacher and he subbed it into a maths program today and the program said that the integration was unsolvable.
so you could try it again, but most likely it's unsolvable.
thanks anyway.
you changed it from 1+tanx (sqaured) / tanx to 1+ u (squared) / u.
so you basically let u = tanx as you subbed it in. however, you initially stated that u=ln(tanx). so...
maybe i'm wrong, but just check it. i dont think its that easy otherwise i woulda got it ages ago. LOL i asked my teacher and he subbed it into a maths program today and the program said that the integration was unsolvable.
so you could try it again, but most likely it's unsolvable.
thanks anyway.
well i got it from a friend, who got it from his teacher. his teacher is this insane asian. funniest guy but also insane smart.
most of his questions are from chinese math competitions but they are usually do-able, if you know how to.
but according to the computer program, this question isn't possible. so don't try it. lol
i wasted 2 days of my life trying.
most of his questions are from chinese math competitions but they are usually do-able, if you know how to.
but according to the computer program, this question isn't possible. so don't try it. lol
i wasted 2 days of my life trying.
pwoh
O_O
When I differentiated that, I got
Wolfram alpha gives me the same derivative...although Online Derivative Calculator gives me 1/ln(tanx) as required.
pwoh
O_O
This is using the derivative above
}}{dx} = \frac{1}{ln(tanx)} - \frac{1}{cosx sinx ln^2(tanx)} \\ \int \frac{d\frac{x}{ln(tanx)}}{dx} = \int \frac{1}{ln(tanx)} - \int \frac{1}{cosx sinx ln^2(tanx)} \\ \int \frac{1}{ln(tanx)} = \frac{x}{ln(tanx)} + \int \frac{1}{cosx sinx ln^2(tanx)} \\\\\\ For \int \frac{1}{cosx sinx ln^2(tanx)}:\\ $Let u$ = ln(tanx)\\ \frac{du}{dx} = \frac{1}{cosx sinx}\\ $Therefore du$ = \frac{dx}{cosx sinx}\\ \int \frac{1}{cosx sinx ln^2(tanx)} = \int \frac{1}{u^2)} = \frac{-1}{u} = \frac{-1}{ln(tan x)} \\\\\\\\ \int \frac{1}{ln(tanx)} = \frac{x}{ln(tanx)} + \int \frac{1}{cosx sinx ln^2(tanx)}\\ \int \frac{1}{ln(tanx)} = \frac{x}{ln(tanx)} + \frac{-1}{ln(tan x)}\\ 2\int \frac{1}{ln(tanx)} = \frac{x}{ln(tanx)}\\ \\ Therefore:\\ \int \frac{1}{ln(tanx)} = \frac{x}{2ln(tanx)}+C)
I haven't checked it though
Edit: Nevermind, it doesn't differentiate to give 1/ln(tan x). Damn
lol
I haven't checked it though
Edit: Nevermind, it doesn't differentiate to give 1/ln(tan x). Damn
lol
Last edited:
theres a button near the bottom that says launch LaTex equation editor
pwoh
O_O
Does that make any difference? o.o
