solve-the-integral-problem-cosx-2-sin2x-dx-

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

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