Integrate √tan^-xdx

∫√<font face="Verdana"><font style="margin: 0px; padding: 0px; box-sizing: inherit; color: rgba(0, 0, 0, 0.54); font-size: 16px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: rgb(255, 255, 255); text-decoration-style: initial; text-decoration-color: initial;"><font size="3"><font style="margin: 0px; padding: 0px; box-sizing: inherit;">(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</font></font></font></font>