0-pi-dx-2-cos-x-

Question Number 140930 by bramlexs22 last updated on 14/May/21

\underset{0}{\int^{π}} \frac{dx}{\sqrt{2} - \cos x}

Answered by Dwaipayan Shikari last updated on 14/May/21

2 \int_{0}^{\infty} \frac{d t}{\sqrt{2} - \frac{1 - t^{2}}{1 + t^{2}}} . \frac{1}{1 + t^{2}} t = t a n \frac{x}{2}

= 2 \int_{0}^{\infty} \frac{d t}{\sqrt{2} t^{2} + \sqrt{2} - 1 + t^{2}} d t = \frac{1}{\sqrt{2} + 1} \int_{- \infty}^{\infty} \frac{d t}{t^{2} + \frac{\sqrt{2} - 1}{\sqrt{2} + 1}} d t

{= \frac{1}{\sqrt{2} + 1} . \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1} [} {t a n}^{- 1} (t \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2} - 1}})]}_{- \infty}^{\infty} d x

= π

Commented by bramlexs22 last updated on 14/May/21

Weirrstrass substitution

Commented by Dwaipayan Shikari last updated on 14/May/21

y e s!

Answered by mathmax by abdo last updated on 14/May/21

I = \int_{0}^{π} \frac{dx}{\sqrt{2} - cosx} we do the changement \tan (\frac{x}{2}) = t \Rightarrow

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

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

= \frac{2}{1 + \sqrt{2}} \int_{0}^{\infty} \frac{dt}{t^{2} + 3 - 2 \sqrt{2}} =_{t = \sqrt{3 - 2 \sqrt{2}} u} \frac{2}{1 + \sqrt{2}} \int_{0}^{\infty} \frac{\sqrt{3 - 2 \sqrt{2}} du}{(3 - 2 \sqrt{2}) (1 + u^{2})}

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

Commented by mathmax by abdo last updated on 14/May/21

but \sqrt{3 - 2 \sqrt{2}} = \sqrt{2} - 1 \Rightarrow I = \frac{π}{(\sqrt{2} - 1) (\sqrt{2} + 1)} = π

Answered by iloveisrael last updated on 14/May/21

R = \underset{0}{\int^{π}} \frac{d x}{a - \cos x}; a > 1

p u t μ = \tan \frac{x}{2}

R = \underset{0}{\int^{π}} \frac{1 + μ^{2}}{a + a μ^{2} - 1 + μ^{2}} . \frac{2}{1 + μ^{2}} d μ

(*) a - 1 + (1 + a) μ^{2}; s e t \frac{a - 1}{a + 1} = p^{2}

a n d μ = p u

R = \frac{2}{a + 1} \underset{0}{\int^{\infty}} \frac{p}{p^{2} (1 + u^{2})} d u

R = \frac{2}{\sqrt{a^{2} - 1}} [\arctan u]_{0}^{\infty}

R = \frac{2}{\sqrt{a^{2} - 1}} . \frac{π}{2} = \frac{π}{\sqrt{a^{2} - 1}}

R = \frac{π}{\sqrt{{(\sqrt{2})}^{2} - 1}} = π