tan-1-1-sinx-1-sinx-dx-

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 = 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 \dots

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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