∫tan^(−1) (√((1−sinx)/(1+sinx))) dx=?


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Question Number 45669 by arvinddayama01@gmail.com last updated on 15/Oct/18

$\int {t a n}^{- 1} \sqrt{\frac{1 - s i n x}{1 + s i n x}} d x = ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
	$t=tan(x/2) now tan^(−1) (√((1−((2t)/(1+t^2 )))/(1+((2t)/(1+t^2 ))))) =(√(((1−t)^2 )/((1+t)^2 ))) =((1−t)/(1+t))=((1−tan(x/2))/(1+tan(x/2)))=tan((π/4)−(x/2)) so tan^(−1) {tan((π/4)−(x/2))}=(π/4)−(x/2) ∫(π/4)−(x/2) dx =((πx)/4)−(x^2 /4)+c$
	$t = t a n \frac{x}{2}$ $n o w {t a n}^{- 1} \sqrt{\frac{1 - \frac{2 t}{1 + t^{2}}}{1 + \frac{2 t}{1 + t^{2}}}}$ $= \sqrt{\frac{{(1 - t)}^{2}}{{(1 + t)}^{2}}} = \frac{1 - t}{1 + t} = \frac{1 - t a n \frac{x}{2}}{1 + t a n \frac{x}{2}} = t a n (\frac{π}{4} - \frac{x}{2})$ $s o {t a n}^{- 1} {t a n (\frac{π}{4} - \frac{x}{2})} = \frac{π}{4} - \frac{x}{2}$ $\int \frac{π}{4} - \frac{x}{2} d x$ $= \frac{π x}{4} - \frac{x^{2}}{4} + c$
	Commented by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18

	$m o s t w e l c o m e . . .$
	Commented by arvinddayama01@gmail.com last updated on 15/Oct/18

	$T h a n k u v e r y m u c h$
	Answered by vermasir last updated on 16/Oct/18

	$\int \tan^{- 1} \sqrt{\frac{1 - c o s (\frac{π}{2} - x)}{1 + c o s (\frac{π}{2} + x)}}$ $\int \tan^{- 1} \sqrt{\frac{2 {s i n}^{2} (\frac{π}{4} - \frac{x}{2})}{2 {c o s}^{2} (\frac{π}{4} - \frac{x}{2})}}$ $\int \tan^{- 1} ∣ t a n (\frac{π}{4} - \frac{x}{2}) ∣ d x$ $\int (\frac{π}{4} - \frac{x}{2}) d x$ $\frac{π}{4} x - \frac{x^{2}}{4} + c$
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