Question

Integrate Root tanx in terms of x
(BIO WAALO KO NAHI AAYEGA )​

Answer 1

Answer:

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

Step-by-step explanation:

∫√(tan x) dx

Let tan x = t²

⇒ sec2 x dx = 2t dt

⇒ dx = [2t / (1 + t⁴)]dt

⇒ Integral  ∫ 2t² / (1 + t⁴) dt

⇒ ∫[(t² + 1) + (t² - 1)] / (1 + t⁴) dt

⇒ ∫(t² + 1) / (1 + t⁴) dt + ∫(t² - 1) / (1 + t⁴) dt

⇒ ∫(1 + 1/t² ) / (t² + 1/t² ) dt + ∫(1 - 1/t² ) / (t² + 1/t² ) dt

⇒ ∫(1 + 1/t² )dt / [(t - 1/t)² + 2] + ∫(1 - 1/t²)dt / [(t + 1/t)² -2]

Let t - 1/t = u for the first integral ⇒ (1 + 1/t² )dt = du

and t + 1/t = v for the 2nd integral ⇒ (1 - 1/t² )dt = dv

Integral

= ∫du/(u² + 2) + ∫dv/(v² - 2)

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

= (1/√2) tan-1 [(t² - 1)/t√2] + (1/2√2) log (t² + 1 - t√2) / t² + 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