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Feb 11, 2015

# Find integral of root tan x .

∫(√tanx)dx

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

Sreejagaths

integral.

Sreejagaths

infinity
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