dx-csc-x-cos-x-

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

\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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