Question

Evaluate: $\int \sqrt{tan \ x} \ dx$

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) tan-1 (u/√2) + (1/2√2) log(v -√2)/(v + √2)l + c 
= (1/√2) tan-1 [(t^2 - 1)/t√2] + (1/2√2) log (t^2 + 1 - t√2) / t^2 + 1 + t√2) + c
 = (1/√2) tan-1 [(tanx - 1)/(√2tan x)] + (1/2√2) log [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c

Hope this helps you !

Answer 2

Answer:

∫√(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) tan-1 (u/√2) + (1/2√2) log(v -√2)/(v + √2)l + c

= (1/√2) tan-1 [(t^2 - 1)/t√2] + (1/2√2) log (t^2 + 1 - t√2) / t^2 + 1 + t√2) + c

= (1/√2) tan-1 [(tanx - 1)/(√2tan x)] + (1/2√2) log [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c

Hope this helps you !