Question

what is the value of this integration root tanx dx ?

Answer 1

∫√(tan x) dx 
Let tan x = t^2 
=> sec^2 x dx = 2t dt 
=> dx = [2t / (1 + t^4)]dt 
=> Integral 
= ∫ 2t^2 / (1 + t^4) dt 
= ∫[(t^2 + 1) + (t^2 - 1)] / (1 + t^4) dt 
= ∫(t^2 + 1) / (1 + t^4) dt + ∫(t^2 - 1) / (1 + t^4) dt 
= ∫(1 + 1/t^2) / (t^2 + 1/t^2) dt + ∫(1 - 1/t^2) / (t^2 + 1/t^2) dt 
= ∫(1 + 1/t^2)dt / [(t - 1/t)^2 + 2] + ∫(1 - 1/t^2)dt / [(t + 1/t)^2 -2] 
Let t - 1/t = u for the first integral => (1 + 1/t^2)dt = du 
and t + 1/t = v for the 2nd integral => (1 - 1/t^2)dt = dv 
Integral 
= ∫du/(u^2 + 2) + ∫dv/(v^2 - 2) 
= (1/√2) arctan (u/√2) + (1/2√2) ln l(v -√2)/(v + √2)l + c 
= (1/√2) arctan [(t^2 - 1)/t√2] + (1/2√2) ln l (t^2 + 1 - t√2) / t^2 + 1 + t√2) + c 
= (1/√2) arctan [(tanx - 1)/(√2tan x)] + (1/2√2) ln l [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c

Hope this helps......