find-0-pi-dx-2-cosx-sinx-2-

Question Number 140976 by Mathspace last updated on 14/May/21

f i n d \int_{0}^{π} \frac{d x}{{(2 - c o s x - s i n x)}^{2}}

Answered by Ar Brandon last updated on 14/May/21

I = \int_{0}^{π} \frac{dx}{{(2 - cosx - sinx)}^{2}}, t = \tan \frac{x}{2}

= \int_{0}^{\infty} \frac{2 dt}{{(2 - \frac{1 - t^{2}}{1 + t^{2}} - \frac{2 t}{1 + t^{2}})}^{2}} \cdot \frac{1}{1 + t^{2}} = 2 \int_{0}^{\infty} \frac{t^{2} + 1}{{({3 t}^{2} - 2 t + 1)}^{2}} dt

= \frac{2}{3} \int_{0}^{\infty} \frac{{3 t}^{2} - 2 t + 1}{{({3 t}^{2} - 2 t + 1)}^{2}} dt + 2 \int_{0}^{\infty} \frac{2 t / 3 + 2 / 3}{{({3 t}^{2} - 2 t + 1)}^{2}} dt

= \frac{2}{3} \int_{0}^{\infty} \frac{dt}{{3 t}^{2} - 2 t + 1} + \frac{4}{3} \int_{0}^{\infty} \frac{t + 1}{{({3 t}^{2} - 2 t + 1)}^{2}} dt

= \frac{2}{3} \int_{0}^{\infty} \frac{dt}{{3 t}^{2} - 2 t + 1} + {[\frac{4}{3} \cdot \frac{t - 1 / 2}{{3 t}^{2} - 2 t + 1} + \frac{4}{3} \int \frac{dt}{{3 t}^{2} - 2 t + 1}]}_{0}^{\infty}

= 2 \int_{0}^{\infty} \frac{dt}{{3 t}^{2} - 2 t + 1} + \frac{2}{3} \cdot {[\frac{2 t - 1}{{3 t}^{2} - 2 t + 1}]}_{0}^{\infty}

= \frac{2}{3} \int_{0}^{\infty} \frac{dt}{t^{2} - \frac{2}{3} t + \frac{1}{3}} + \frac{2}{3} = \frac{2}{3} \int_{0}^{\infty} \frac{dt}{{(t - \frac{1}{3})}^{2} + \frac{2}{9}} + \frac{2}{3}

= \frac{2}{3} \cdot {[\frac{3}{\sqrt{2}} \arctan (\frac{3 t - 1}{\sqrt{2}})]}_{0}^{\infty} + \frac{2}{3} = \sqrt{2} (\frac{π}{2} + \tan^{- 1} (\frac{\sqrt{2}}{2})) + \frac{2}{3}

\approx 3.758527887

Commented by Ar Brandon last updated on 14/May/21

Mr {MJS}^{'} Ostrogradsky (line 4 \to line 5)

Commented by Ar Brandon last updated on 14/May/21

Commented by Tawa11 last updated on 06/Nov/21

Weldone sir

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