∫ (dx/(a+bcosx)) ∫ (dx/(a−bsinx))


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Question Number 176648 by mokys last updated on 23/Sep/22

$\int \frac{d x}{a + b c o s x}$ $\int \frac{d x}{a - b s i n x}$
Commented by mokys last updated on 24/Sep/22

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	Answered by Ar Brandon last updated on 24/Sep/22

	$\int \frac{d x}{a + b \cos x} = \int \frac{2}{a + b (\frac{1 - t^{2}}{1 + t^{2}})} \cdot \frac{1}{1 + t^{2}} d t$ $= 2 \int \frac{d t}{(a - b) t^{2} + a + b}$ $= \frac{2}{\sqrt{a^{2} - b^{2}}} \arctan (t \sqrt{\frac{a - b}{a + b}}) + C$ $= \frac{2}{\sqrt{a^{2} - b^{2}}} \arctan (\sqrt{\frac{a - b}{a + b}} \tan (\frac{x}{2})) + C$
	Answered by BaliramKumar last updated on 24/Sep/22

	$\int \frac{d x}{a - b s i n x} = \int \frac{d x}{a - b [\frac{2 t a n (\frac{x}{2})}{1 + {t a n}^{2} (\frac{x}{2})}]}$ $L e t t a n (\frac{x}{2}) = y t h e n {s e c}^{2} (\frac{x}{2}) \cdot \frac{1}{2} \cdot d x = d y$ $[1 + {t a n}^{2} (\frac{x}{2})] \cdot \frac{d x}{2} = d y, d x = \frac{2 d y}{1 + y^{2}}$ $\int \frac{2}{a - b (\frac{2 y}{1 + y^{2}})} \cdot \frac{d y}{1 + y^{2}} = \int \frac{2 d y}{a + {a y}^{2} - 2 b y}$ $\frac{2}{a} \int \frac{d y}{y^{2} - 2 (\frac{b}{a}) y + 1} = \frac{2}{a} \int \frac{d y}{{(y - \frac{b}{a})}^{2} + {(\frac{\sqrt{a^{2} - b^{2}}}{a})}^{2}}$ $\frac{2}{a} \cdot \frac{a}{\sqrt{a^{2} - b^{2}}} \cdot {t a n}^{- 1} (\frac{a y - b}{\sqrt{a^{2} - b^{2}}}) + C$ $\frac{2}{\sqrt{a^{2} - b^{2}}} \cdot {t a n}^{- 1} [\frac{a \cdot t a n (\frac{x}{2}) - b}{\sqrt{a^{2} - b^{2}}}] + C$
	Commented by Tawa11 last updated on 25/Sep/22

	$Great sir$
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