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Asked by Adarsh Udayan

Feb 11, 2015

Find integral of root tan x .

∫(√tanx)dx

Answers(4)

Answer

Raghunath Reddy

Member since Apr 11, 2014

∫√(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

Syeda

Member since Jan 25, 2017

Answer.


∫√(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

Hhh

integral.

Hhh

infinity
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