Question-167301

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

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}) \dots .