solve the integral problem ∫((cosx)/(2−sin2x))dx


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Question Number 111616 by mathdave last updated on 04/Sep/20

$s o l v e t h e i n t e g r a l p r o b l e m$ $\int \frac{\cos x}{2 - \sin 2 x} d x$
	Answered by Her_Majesty last updated on 05/Sep/20

	$\int \frac{\cos x}{2 - \sin 2 x} d x =$ $(use t = \tan \frac{x}{2})$ $= - \int \frac{t^{2} - 1}{t^{4} + 2 t^{3} + 2 t^{2} - 2 t + 1} d t =$ $= \frac{\sqrt{3}}{6} \int (\frac{t - 1}{t^{2} + (1 - \sqrt{3}) t + 2 - \sqrt{3}} - \frac{t - 1}{t^{2} + (1 + \sqrt{3}) t + 2 + \sqrt{3}}) d t =$ $= \frac{\sqrt{3}}{12} \ln \frac{t^{2} + (1 - \sqrt{3}) t + 2 - \sqrt{3}}{t^{2} + (1 + \sqrt{3}) t + 2 + \sqrt{3}} - \frac{1}{2} (\arctan ((1 - \sqrt{3}) t - 1) + \arctan ((1 + \sqrt{3}) t - 1))$ $now put t = \tan \frac{x}{2}$
	Answered by ajfour last updated on 05/Sep/20
	$let x=(π/4)−θ I=−∫((cos ((π/4)−θ)dθ)/(2−cos 2θ)) =−(1/(2(√2)))∫((cos θ+sin θ)/(sin^2 θ))dθ =(1/(2(√2))){cosec θ−ln ∣cosec θ−cot θ∣}+c I =(1/(2(√2)sin ((π/4)−x)))−(1/(2(√2)))ln∣tan ((π/8)−(x/2))∣+c$
	$l e t x = \frac{π}{4} - θ$ $I = - \int \frac{\cos (\frac{π}{4} - θ) d θ}{2 - \cos 2 θ}$ $= - \frac{1}{2 \sqrt{2}} \int \frac{\cos θ + \sin θ}{\sin^{2} θ} d θ$ $= \frac{1}{2 \sqrt{2}} {cosec θ - \ln ∣ cosec θ - \cot θ ∣} + c$ $I = \frac{1}{2 \sqrt{2} \sin (\frac{π}{4} - x)} - \frac{1}{2 \sqrt{2}} \ln ∣ \tan (\frac{π}{8} - \frac{x}{2}) ∣ + c$
	Commented by Her_Majesty last updated on 05/Sep/20

	$g r e a t, I {h a v e n}^{'} t t h o u g h t o f t h i s$
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