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Question Number 167301 by mnjuly1970 last updated on 12/Mar/22

	Answered by amin96 last updated on 12/Mar/22

	$I = \int_{0}^{\frac{π}{8}} \frac{1}{\sqrt{2} (1 - \cos (\frac{π}{4} - x))} dx \frac{π}{4} - x = t$ $t [\frac{π}{8}; \frac{π}{4}] I = \frac{1}{\sqrt{2}} \int_{\frac{π}{8}}^{\frac{π}{4}} \frac{dt}{(1 - \cos (t))} = \frac{1}{2 \sqrt{2}} \int_{\frac{π}{8}}^{\frac{π}{4}} \csc^{2} (\frac{t}{2}) dt =$ $= - \frac{1}{\sqrt{2}} {[\cot (\frac{t}{2})]}_{\frac{π}{8}}^{\frac{π}{4}} = - \frac{1}{\sqrt{2}} [\frac{\sin (\frac{π}{16} - \frac{π}{8})}{\sin (\frac{π}{8}) \sin (\frac{π}{16})}] =$ $= \frac{1}{\sqrt{2}} [\frac{1}{\sin (\frac{π}{8})}] = \frac{\sqrt{2}}{2 \sqrt{\frac{1 - \cos (\frac{π}{4}}{2}}} = \frac{1}{\sqrt{1 - \frac{\sqrt{2}}{2}}} =$ $= \frac{\sqrt{2}}{\sqrt{2 - \sqrt{2}}} = \frac{\sqrt{2} \times \sqrt{2 + \sqrt{2}}}{\sqrt{2}} = \sqrt{2 + \sqrt{2}} by M . A$
	Commented by mnjuly1970 last updated on 12/Mar/22

	$t h a n k s a l o t s i r$
	Answered by MJS_new last updated on 12/Mar/22

	$α = β = 2$
	Answered by Mathspace last updated on 13/Mar/22
	$changement tan((x/2))=t give I=∫_0 ^(tan((π/(16)))) ((2dt)/((1+t^2 )((√2)−((2t)/(1+t^2 ))−((1−t^2 )/(1+t^2 ))))) =2∫_0 ^(tan((π/(16)))) (dt/( (√2)(1+t^2 )−2t−1+t^2 )) =2∫_0 ^(tan((π/(16)))) (dt/( (√2)+(1+(√2))t^2 −2t−1)) =2∫_0 ^(tan((π/(16)))) (dt/((1+(√2))t^2 −2t+(√2)−1)) roots! Δ^′ =1−((√2)+1)((√2)−1) =1−(2−1)=1−1=0 one root x_0 =(1/(1+(√2)))=(√2)−1 ⇒ (1+(√2))t^2 −2t+(√2)−1=(1+(√2))(t−(√2)+1)^2 ⇒I=2∫_0 ^(tan((π/(16)))) (dt/((1+(√2))(t−(√2)+1)^2 )) =2((√2)−1)∫_0 ^(tan((π/(16)))) (dt/((t−(√2)+1)^2 )) =−2((√2)−1)[(1/(t−(√2)+1))]_0 ^(tan((π/(16)))) =2(1−(√2)){(1/(tan((π/(16)))−(√2)+1))−(1/(1+(√2)))} rest to calculate tan((π/(16))) by the formulae tan(2x)=((2tanx)/(1−tan^2 x)) ⇒ tan((π/8))=(√2)−1=((2x)/(1−x^2 )) x=tan((π/(16)))....$
	$c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int_{0}^{t a n (\frac{π}{16})} \frac{2 d t}{(1 + t^{2}) (\sqrt{2} - \frac{2 t}{1 + t^{2}} - \frac{1 - t^{2}}{1 + t^{2}})}$ $= 2 \int_{0}^{t a n (\frac{π}{16})} \frac{d t}{\sqrt{2} (1 + t^{2}) - 2 t - 1 + t^{2}}$ $= 2 \int_{0}^{t a n (\frac{π}{16})} \frac{d t}{\sqrt{2} + (1 + \sqrt{2}) t^{2} - 2 t - 1}$ $= 2 \int_{0}^{t a n (\frac{π}{16})} \frac{d t}{(1 + \sqrt{2}) t^{2} - 2 t + \sqrt{2} - 1}$ $r o o t s!$ $Δ^{^{'}} = 1 - (\sqrt{2} + 1) (\sqrt{2} - 1)$ $= 1 - (2 - 1) = 1 - 1 = 0$ $o n e r o o t x_{0} = \frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1 \Rightarrow$ $(1 + \sqrt{2}) t^{2} - 2 t + \sqrt{2} - 1 = (1 + \sqrt{2}) {(t - \sqrt{2} + 1)}^{2}$ $\Rightarrow I = 2 \int_{0}^{t a n (\frac{π}{16})} \frac{d t}{(1 + \sqrt{2}) {(t - \sqrt{2} + 1)}^{2}}$ $= 2 (\sqrt{2} - 1) \int_{0}^{t a n (\frac{π}{16})} \frac{d t}{{(t - \sqrt{2} + 1)}^{2}}$ $= - 2 (\sqrt{2} - 1) {[\frac{1}{t - \sqrt{2} + 1}]}_{0}^{t a n (\frac{π}{16})}$ $= 2 (1 - \sqrt{2}) {\frac{1}{t a n (\frac{π}{16}) - \sqrt{2} + 1} - \frac{1}{1 + \sqrt{2}}}$ $r e s t t o c a l c u l a t e t a n (\frac{π}{16}) b y t h e f o r m u l a e$ $t a n (2 x) = \frac{2 t a n x}{1 - {t a n}^{2} x} \Rightarrow$ $t a n (\frac{π}{8}) = \sqrt{2} - 1 = \frac{2 x}{1 - x^{2}}$ $x = t a n (\frac{π}{16}) . . . .$
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