calculate-0-pi-2-xcosx-cos-2x-dx-

Question Number 120316 by Bird last updated on 30/Oct/20

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{x c o s x}{c o s (2 x)} d x

Answered by mathmax by abdo last updated on 30/Oct/20

let I = \int_{0}^{\frac{π}{2}} \frac{x}{cosx + sinx} dx we have proved that I = \frac{π}{4 \sqrt{2}} \ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})

changement x = t - \frac{π}{2} give

I = - \int_{0}^{\frac{π}{2}} \frac{t - \frac{π}{2}}{sint - cost} dt = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{dt}{sint - cost} - \int_{0}^{\frac{π}{2}} \frac{t}{sint - cost}

= \int_{0}^{\frac{π}{2}} \frac{x}{cosx - sinx} dx + \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{dx}{sinx - cosx} \Rightarrow

2 I = \int_{0}^{\frac{π}{2}} x (\frac{1}{cosx + sinx} + \frac{1}{cosx - sinx}) dx + \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{dx}{sinx - cosx}

= \int_{0}^{\frac{π}{2}} \frac{x (2 cosx)}{\cos^{2} x - \sin^{2} x} dx + \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{dx}{sinx - cosx}

\Rightarrow 2 \int_{0}^{\frac{π}{2}} \frac{xcosx}{\cos (2 x)} dx = 2 I - \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{dx}{sinx - cosx} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{xcosx}{\cos (2 x)} dx = I - \frac{π}{4} \int_{0}^{\frac{π}{2}} \frac{dx}{sinx - cosx} we have

\int_{0}^{\frac{π}{2}} \frac{dx}{sinx - cosx} =_{\tan (\frac{x}{2}) = t} \int_{0}^{1} \frac{2 dt}{(1 + t^{2}) (\frac{2 t}{1 + t^{2}} - \frac{1 - t^{2}}{1 + t^{2}})}

= \int_{0}^{1} \frac{2 dt}{2 t - 1 + t^{2}} = 2 \int_{0}^{1} \frac{dt}{t^{2} + 2 t - 1} = 2 \int_{0}^{1} \frac{dt}{{(t + 1)}^{2} - 2}

= 2 \int_{0}^{1} \frac{dt}{(t + 1 - \sqrt{2}) (t + 1 + \sqrt{2})} = \frac{2}{2 \sqrt{2}} \int_{0}^{1} (\frac{1}{t + 1 - \sqrt{2}} - \frac{1}{t + 1 + \sqrt{2}})

= \frac{1}{\sqrt{2}} {[\ln ∣ \frac{t + 1 - \sqrt{2}}{t + 1 + \sqrt{2}} ∣]}_{0}^{1} = \frac{1}{\sqrt{2}} \ln ∣ \frac{2 - \sqrt{2}}{2 + \sqrt{2}} ∣ - \ln ∣ \frac{1 - \sqrt{2}}{1 + \sqrt{2}} ∣

= \frac{1}{\sqrt{2}} \ln (\frac{2 - \sqrt{2}}{2 + \sqrt{2}}) - \ln (\frac{\sqrt{2} - 1}{\sqrt{2} + 1}) \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{xcosx}{\cos (2 x)} dx = \frac{π}{4 \sqrt{2}} \ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1}) - \frac{π}{4 \sqrt{2}} \ln (\frac{2 - \sqrt{2}}{2 + \sqrt{2}}) + \frac{π}{4} \ln (\frac{\sqrt{2} - 1}{\sqrt{2} + 1})

= \frac{π}{4} (\frac{1}{\sqrt{2}} - 1) \ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1}) - \frac{π}{4 \sqrt{2}} \ln (\frac{2 - \sqrt{2}}{2 + \sqrt{2}})