B=∫ (√((sin 2x−1)/(cos 2x−1))) dx =?


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Question Number 166488 by bobhans last updated on 21/Feb/22

$B = \int \sqrt{\frac{\sin 2 x - 1}{\cos 2 x - 1}} dx = ?$
	Answered by greogoury55 last updated on 21/Feb/22
	$B=∫ (√((1−sin 2x)/(1−cos 2x))) dx B=∫ (√((sin^2 x+cos^2 x−2sin x cos x)/(2sin^2 x))) dx B=(1/( (√2))) ∫ ((∣sin x−cos x∣)/(∣sin x∣)) dx → { ((∣sin x−cos x∣= { ((sin x−cos x)),((cos x−sin x)) :})),((∣sin x∣= { ((sin x)),((−sin x)) :})) :} B_1 =(1/( (√2)))∫(1−cot x)dx=(1/( (√2)))(x−ln ∣sin x∣)+c B_2 =(1/( (√2)))∫(−1+cot x)dx=(1/( (√2)))(−x+ln ∣sin x∣)+c$
	$B = \int \sqrt{\frac{1 - \sin 2 x}{1 - \cos 2 x}} d x$ $B = \int \sqrt{\frac{\sin^{2} x + \cos^{2} x - 2 \sin x \cos x}{2 \sin^{2} x}} d x$ $B = \frac{1}{\sqrt{2}} \int \frac{∣ \sin x - \cos x ∣}{∣ \sin x ∣} d x$ $\to {\begin{cases} ∣ \sin x - \cos x ∣= {\begin{cases} \sin x - \cos x \\ \cos x - \sin x \end{cases} \\ ∣ \sin x ∣= {\begin{cases} \sin x \\ - \sin x \end{cases} \end{cases}$ $B_{1} = \frac{1}{\sqrt{2}} \int (1 - \cot x) d x = \frac{1}{\sqrt{2}} (x - \ln ∣ \sin x ∣) + c$ $B_{2} = \frac{1}{\sqrt{2}} \int (- 1 + \cot x) d x = \frac{1}{\sqrt{2}} (- x + \ln ∣ \sin x ∣) + c$
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