∫(x/(sin^2 x−3))dx=?


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Question Number 102198 by Study last updated on 07/Jul/20

$\int \frac{x}{{s i n}^{2} x - 3} d x = ?$
	Answered by mathmax by abdo last updated on 08/Jul/20

	$not resoluble \sqrt{!}$
	Answered by Dwaipayan Shikari last updated on 08/Jul/20

	$\int \frac{x}{(s i n x + \sqrt{3}) (s i n x - \sqrt{3})}$ $\frac{x}{2 \sqrt{3}} \int \frac{1}{s i n x - \sqrt{3}} - \frac{1}{s i n x - \sqrt{3}} d x - \int (\int \frac{1}{{s i n}^{2} x - 3}) d x$ $x \frac{1}{2 \sqrt{3}} (I_{a} - I_{b}) - \int (\int \frac{1}{{s i n}^{2} x - 3}) d x$ $I_{a} = \int \frac{1}{s i n x - \sqrt{3}} d x$ $= 2 \int \frac{d t}{(\frac{2 t}{t^{2} + 1} - \sqrt{3}) (t^{2} + 1)} = 2 \int \frac{1}{2 t - \sqrt{3} t^{2} - \sqrt{3}} d t = - \frac{2}{\sqrt{3}} \int \frac{1}{t^{2} - \frac{2 t}{\sqrt{3}} + 1}$ $= - \frac{2}{\sqrt{3}} \int \frac{1}{{(t - \frac{1}{\sqrt{3}})}^{2} + {(\sqrt{\frac{2}{3}})}^{2}} d t$ $= - \frac{1}{\sqrt{2}} {t a n}^{- 1} \frac{\sqrt{3} t - 1}{\sqrt{2}}$ $= - \frac{1}{\sqrt{2}} {t a n}^{- 1} \frac{\sqrt{3} t a n \frac{x}{2} - 1}{\sqrt{2}}$ $I_{b} = \int \frac{1}{s i n x + \sqrt{3}} = 2 \int \frac{1}{(\frac{2 t}{t^{2} + 1} + \sqrt{3}) (t^{2} + 1)} d t$ $= \frac{2}{\sqrt{3}} \int \frac{1}{t^{2} + \frac{2 t}{\sqrt{3}} + 1} = \frac{1}{\sqrt{2}} {t a n}^{- 1} \frac{\sqrt{3} t a n \frac{x}{2} + 1}{\sqrt{2}}$ $\frac{x}{2 \sqrt{3}} (I_{a} - I_{b}) = - \frac{x}{2 \sqrt{6}} ({t a n}^{- 1} \frac{\sqrt{3} t a n \frac{x}{2} + 1}{\sqrt{2}} + {t a n}^{- 1} \frac{\sqrt{3} t a n \frac{x}{2} - 1}{\sqrt{2}}) . . . .$ $. . . . . c o n t i n u e$ $B u t n o w w e h a v e t o i n t e g r a t e$ $\int (I_{a} - I_{b}) = - \int {t a n}^{- 1} (\frac{\frac{2 \sqrt{3} t a n \frac{x}{2}}{\sqrt{2}}}{1 - \frac{3 {t a n}^{2} \frac{x}{2} - 1}{2}})$ $- \int {t a n}^{- 1} (\frac{2 \sqrt{6} t a n \frac{x}{2}}{1 - 3 {t a n}^{2} \frac{x}{2}}) . . . . . . . . c o n t i n u e$ $= - {x t a n}^{- 1} (\frac{2 \sqrt{6} t a n \frac{x}{2}}{1 - 3 {t a n}^{2} \frac{x}{2}}) + \int \frac{x \sqrt{6} {s e c}^{2} \frac{x}{2}}{1 + 9 {t a n}^{4} \frac{x}{2} + 18 {t a n}^{2} \frac{x}{2}}$ $\frac{x}{- 2 \sqrt{3}} (I_{a} - I_{b}) - \int (I_{a} - I_{b})$ $= - \frac{(x + 1)}{2 \sqrt{6}} ({t a n}^{- 1} (\frac{2 \sqrt{6} t a n \frac{x}{2}}{1 - 3 {t a n}^{2} \frac{x}{2}})) + \int \frac{x \sqrt{6} {s e c}^{2} \frac{x}{2}}{1 + 9 {t a n}^{4} \frac{x}{2} + 18 {t a n}^{2} \frac{x}{2}} + C o n s t a n t$
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