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Feb 11, 2015

# Find integral of root tan x .

âˆ«(âˆštanx)dx

Raghunath Reddy

Member since Apr 11, 2014

âˆ«âˆš(tan x) dx Let tan x = t2 â‡’ sec2 x dx = 2t dt â‡’ dx = [2t / (1 + t4)]dt â‡’ Integral  âˆ« 2t2 / (1 + t4) dt â‡’ âˆ«[(t2 + 1) + (t2 - 1)] / (1 + t4) dt â‡’ âˆ«(t2 + 1) / (1 + t4) dt + âˆ«(t2 - 1) / (1 + t4) dt â‡’ âˆ«(1 + 1/t2 ) / (t2 + 1/t2 ) dt + âˆ«(1 - 1/t2 ) / (t2 + 1/t2 ) dt â‡’ âˆ«(1 + 1/t2 )dt / [(t - 1/t)2 + 2] + âˆ«(1 - 1/t2)dt / [(t + 1/t)2 -2] Let t - 1/t = u for the first integral â‡’ (1 + 1/t2 )dt = du and t + 1/t = v for the 2nd integral â‡’ (1 - 1/t2 )dt = dv Integral = âˆ«du/(u2 + 2) + âˆ«dv/(v2 - 2) = (1/âˆš2) tan-1 (u/âˆš2) + (1/2âˆš2) log(v -âˆš2)/(v + âˆš2)l + c = (1/âˆš2) tan-1 [(t2 - 1)/tâˆš2] + (1/2âˆš2) log (t2 + 1 - tâˆš2) / t2 + 1 + tâˆš2) + c = (1/âˆš2) tan-1 [(tanx - 1)/(âˆš2tan x)] + (1/2âˆš2) log [tanx + 1 - âˆš(2tan x)] / [tan x + 1 + âˆš(2tan x)] + c

Syeda

Member since Jan 25, 2017

Answer. âˆ«âˆš(tan x) dx Let tan x = t2 â‡’ sec2 x dx = 2t dt â‡’ dx = [2t / (1 + t4)]dt â‡’ Integral  âˆ« 2t2 / (1 + t4) dt â‡’ âˆ«[(t2 + 1) + (t2 - 1)] / (1 + t4) dt â‡’ âˆ«(t2 + 1) / (1 + t4) dt + âˆ«(t2 - 1) / (1 + t4) dt â‡’ âˆ«(1 + 1/t2 ) / (t2 + 1/t2 ) dt + âˆ«(1 - 1/t2 ) / (t2 + 1/t2 ) dt â‡’ âˆ«(1 + 1/t2 )dt / [(t - 1/t)2 + 2] + âˆ«(1 - 1/t2)dt / [(t + 1/t)2 -2] Let t - 1/t = u for the first integral â‡’ (1 + 1/t2 )dt = du and t + 1/t = v for the 2nd integral â‡’ (1 - 1/t2 )dt = dv Integral = âˆ«du/(u2 + 2) + âˆ«dv/(v2 - 2) = (1/âˆš2) tan-1 (u/âˆš2) + (1/2âˆš2) log(v -âˆš2)/(v + âˆš2)l + c = (1/âˆš2) tan-1 [(t2 - 1)/tâˆš2] + (1/2âˆš2) log (t2 + 1 - tâˆš2) / t2 + 1 + tâˆš2) + c = (1/âˆš2) tan-1 [(tanx - 1)/(âˆš2tan x)] + (1/2âˆš2) log [tanx + 1 - âˆš(2tan x)] / [tan x + 1 + âˆš(2tan x)] + c

Hhh

integral.

Hhh

infinity
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