∫ ((√(tan x))/( (√(tan^2 x − 1)))) dx


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Question Number 127368 by I want to learn more last updated on 29/Dec/20

$\int \frac{\sqrt{\tan x}}{\sqrt{\tan^{2} x - 1}} dx$
	Answered by liberty last updated on 29/Dec/20
	$let → { ((tan x ≥0)),((tan^2 x > 1)) :} L = ∫ (√(((sin x)/(cos x))/((sin^2 x−cos^2 x)/(cos^2 x)))) dx L = ∫ (√((sin x cos x)/(sin^2 x−cos^2 x))) dx =(1/( (√2)))∫ (√(−tan 2x)) dx L = (1/( (√2))) ∫ (√(tan (−2x))) dx putting z = −2x ⇒L = −((√2)/4) ∫ (√(tan z)) dz L = −((√2)/4) ( (1/( (√2))) tan^(−1) (((tan z−1)/( (√(2tan z)))))+((√2)/4)ln ∣ ((tan z−(√(2tan z)) +1)/(tan z+(√(2tan z)) +1)) ∣ + c L=(1/4)tan^(−1) (((tan 2x+1)/( (√(−2tan 2x)))))+(1/8)ln ∣((tan 2x+(√(−2tan 2x))−1)/(tan 2x−(√(−2tan 2x))−1)) ∣ + c$
	$l e t \to {\begin{cases} \tan x ⩾ 0 \\ \tan^{2} x > 1 \end{cases}$ $L = \int \sqrt{\frac{\frac{\sin x}{\cos x}}{\frac{\sin^{2} x - \cos^{2} x}{\cos^{2} x}}} d x$ $L = \int \sqrt{\frac{\sin x \cos x}{\sin^{2} x - \cos^{2} x}} d x = \frac{1}{\sqrt{2}} \int \sqrt{- \tan 2 x} d x$ $L = \frac{1}{\sqrt{2}} \int \sqrt{\tan (- 2 x)} d x$ $p u t t i n g z = - 2 x \Rightarrow L = - \frac{\sqrt{2}}{4} \int \sqrt{\tan z} d z$ $L = - \frac{\sqrt{2}}{4} (\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan z - 1}{\sqrt{2 \tan z}}) + \frac{\sqrt{2}}{4} \ln ∣ \frac{\tan z - \sqrt{2 \tan z} + 1}{\tan z + \sqrt{2 \tan z} + 1} ∣ + c$ $L = \frac{1}{4} \tan^{- 1} (\frac{\tan 2 x + 1}{\sqrt{- 2 \tan 2 x}}) + \frac{1}{8} \ln ∣ \frac{\tan 2 x + \sqrt{- 2 \tan 2 x} - 1}{\tan 2 x - \sqrt{- 2 \tan 2 x} - 1} ∣ + c$
	Commented by I want to learn more last updated on 29/Dec/20

	$Thanks sir, i appreciate .$
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