∫ (dx/(csc x+ cos x)) =?


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Question Number 175602 by cortano1 last updated on 03/Sep/22

$\int \frac{dx}{\csc x + \cos x} = ?$
Commented by infinityaction last updated on 04/Sep/22
$∫((2sinx )/(2+2sinx.cosx ))dx I =∫_Ψ (((sinx+cosx) dx)/( 3−(sinx−cosx)^2 ))+∫_Φ (((sinx −cosx) dx)/(1+(sinx+cosx)^2 )) sinx − cosx = y (sinx + cosx)dx = dy Ψ = ∫ (dx/( ((√3))^2 −y^2 )) = (1/(2(√3)))log∣(((√3)+y)/( (√3)−y))∣+c_1 Ψ = (1/(2(√3)))log∣(((√3)+sinx−cosx )/( (√3)−sinx+cosx ))∣ now Φ = ∫(((sinx − cosx)dx )/(1+(sinx+cosx)^2 )) sinx + cosx = r −(sinx − cosx )dx = dr Φ = −∫(dr/(1+r^2 )) = −tan^(−1) r +c_2 Φ = −{tan^(−1) (sinx + cosx)}+c_2 I = Ψ+Φ I = (1/(2(√3)))log∣(((√3) +sinx−cosx )/( (√3) −sinx+cosx ))∣− {tan^(−1) (sinx+cosx)}+λ$
$\int \frac{2 \sin x}{2 + 2 \sin x . \cos x} d x$ $I = \int_{Ψ} \frac{(\sin x + \cos x) d x}{3 - {(\sin x - \cos x)}^{2}} + \int_{Φ} \frac{(\sin x - \cos x) d x}{1 + {(\sin x + \cos x)}^{2}}$ $\sin x - \cos x = y$ $(\sin x + \cos x) d x = d y$ $Ψ = \int \frac{d x}{{(\sqrt{3})}^{2} - y^{2}} = \frac{1}{2 \sqrt{3}} \log ∣ \frac{\sqrt{3} + y}{\sqrt{3} - y} ∣ + c_{1}$ $Ψ = \frac{1}{2 \sqrt{3}} \log ∣ \frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - \sin x + \cos x} ∣$ $n o w$ $Φ = \int \frac{(\sin x - \cos x) d x}{1 + {(\sin x + \cos x)}^{2}}$ $\sin x + \cos x = r$ $- (\sin x - \cos x) d x = d r$ $Φ = - \int \frac{d r}{1 + r^{2}} = - \tan^{- 1} r + c_{2}$ $Φ = - {\tan^{- 1} (\sin x + \cos x)} + c_{2}$ $I = Ψ + Φ$ $I = \frac{1}{2 \sqrt{3}} \log ∣ \frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - \sin x + \cos x} ∣ -$ ${\tan^{- 1} (\sin x + \cos x)} + λ$
Commented by Tawa11 last updated on 15/Sep/22

$Great sir .$
	Answered by Frix last updated on 03/Sep/22

	$\int \frac{d x}{\csc x + \cos x} \underset{d x = \frac{2 d t}{t^{2} + 1}}{\overset{t = \tan (\frac{x}{2} + \frac{π}{8})}{=}}$ $= 2 \sqrt{2} \int \frac{t^{2} + 2 t - 1}{t^{4} + 10 t^{2} + 1} d t =$ $= \frac{\sqrt{3}}{3} \int (\frac{t - 3 + \sqrt{6}}{t^{2} + 5 - 2 \sqrt{6}} - \frac{t - 3 - \sqrt{6}}{t^{2} + 5 + 2 \sqrt{6}}) d t =$ $= \frac{\sqrt{3}}{6} \ln (t^{2} + 5 - 2 \sqrt{6}) - \tan^{- 1} ((\sqrt{3} + \sqrt{2}) t) -$ $- \frac{\sqrt{3}}{6} \ln (t^{2} + 5 + 2 \sqrt{6}) + \tan^{- 1} ((\sqrt{3} - \sqrt{2}) t) =$ $= \frac{\sqrt{3}}{6} \ln \frac{t^{2} + 5 - 2 \sqrt{6}}{t^{2} + 5 + 2 \sqrt{6}} - \tan^{- 1} \frac{2 \sqrt{2} t}{t^{2} + 1}$ $the rest is easy$
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