hello abhishek

finding integral of root tanx

∫√(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

if you still are unable to understand the method, you can also refer to this video: https://www.youtube.com/watch?v=CcP7YKr5qBo

hope this helps