find ∫_0 ^π (dx/((2−cosx−sinx)^2 ))


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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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