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Asked by Adarsh Udayan

Feb 11, 2015

âˆ«(âˆštanx)dx

Raghunath Reddy

Member since Apr 11, 2014

âˆ«âˆš(tan x) dx

Let tan x = t^{2}

â‡’ sec^{2} x dx = 2t dt

â‡’ dx = [2t / (1 + t^{4})]dt

â‡’ Integral âˆ« 2t^{2} / (1 + t^{4}) dt

â‡’ âˆ«[(t^{2} + 1) + (t^{2} - 1)] / (1 + t^{4}) dt

â‡’ âˆ«(t^{2} + 1) / (1 + t^{4}) dt + âˆ«(t^{2} - 1) / (1 + t^{4}) dt

â‡’ âˆ«(1 + 1/t^{2} ) / (t^{2} + 1/t^{2} ) dt + âˆ«(1 - 1/t^{2} ) / (t^{2} + 1/t^{2} ) dt

â‡’ âˆ«(1 + 1/t^{2} )dt / [(t - 1/t)^{2} + 2] + âˆ«(1 - 1/t^{2})dt / [(t + 1/t)^{2} -2]

Let t - 1/t = u for the first integral â‡’ (1 + 1/t^{2} )dt = du

and t + 1/t = v for the 2nd integral â‡’ (1 - 1/t^{2} )dt = dv

Integral

= âˆ«du/(u^{2} + 2) + âˆ«dv/(v^{2} - 2)

= (1/âˆš2) tan^{-1} (u/âˆš2) + (1/2âˆš2) log(v -âˆš2)/(v + âˆš2)l + c

= (1/âˆš2) tan^{-1} [(t^{2} - 1)/tâˆš2] + (1/2âˆš2) log (t^{2} + 1 - tâˆš2) / t^{2} + 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 = t^{2}

â‡’ sec^{2} x dx = 2t dt

â‡’ dx = [2t / (1 + t^{4})]dt

â‡’ Integral âˆ« 2t^{2} / (1 + t^{4}) dt

â‡’ âˆ«[(t^{2} + 1) + (t^{2} - 1)] / (1 + t^{4}) dt

â‡’ âˆ«(t^{2} + 1) / (1 + t^{4}) dt + âˆ«(t^{2} - 1) / (1 + t^{4}) dt

â‡’ âˆ«(1 + 1/t^{2} ) / (t^{2} + 1/t^{2} ) dt + âˆ«(1 - 1/t^{2} ) / (t^{2} + 1/t^{2} ) dt

â‡’ âˆ«(1 + 1/t^{2} )dt / [(t - 1/t)^{2} + 2] + âˆ«(1 - 1/t^{2})dt / [(t + 1/t)^{2} -2]

Let t - 1/t = u for the first integral â‡’ (1 + 1/t^{2} )dt = du

and t + 1/t = v for the 2nd integral â‡’ (1 - 1/t^{2} )dt = dv

Integral

= âˆ«du/(u^{2} + 2) + âˆ«dv/(v^{2} - 2)

= (1/âˆš2) tan^{-1} (u/âˆš2) + (1/2âˆš2) log(v -âˆš2)/(v + âˆš2)l + c

= (1/âˆš2) tan^{-1} [(t^{2} - 1)/tâˆš2] + (1/2âˆš2) log (t^{2} + 1 - tâˆš2) / t^{2} + 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

âˆ«âˆš(tan x) dx

Let tan x = t

â‡’ sec

â‡’ dx = [2t / (1 + t

â‡’ Integral âˆ« 2t

â‡’ âˆ«[(t

â‡’ âˆ«(t

â‡’ âˆ«(1 + 1/t

â‡’ âˆ«(1 + 1/t

Let t - 1/t = u for the first integral â‡’ (1 + 1/t

and t + 1/t = v for the 2nd integral â‡’ (1 - 1/t

Integral

= âˆ«du/(u

= (1/âˆš2) tan

= (1/âˆš2) tan

= (1/âˆš2) tan

Hhh

integral.

Hhh

infinity