0-pi-x-2-cos-2x-dx-0-Prove-or-Disprove-

Question Number 126865 by Lordose last updated on 24/Dec/20

\int_{0}^{π} \frac{x}{2 + \cos (2 x)} dx = 0

Prove or Disprove

Commented by Ar Brandon last updated on 25/Dec/20

Commented by bramlexs22 last updated on 25/Dec/20

Commented by liberty last updated on 25/Dec/20

Commented by MJS_new last updated on 25/Dec/20

first, plot f (x) = \frac{x}{2 + \cos 2 x} for 0 ⩽ x ⩽ π to see

the integral obviously is > 0

Answered by Ar Brandon last updated on 25/Dec/20

I = \int_{0}^{π} \frac{x}{2 + \cos (2 x)} dx

I = \int_{0}^{π} \frac{π - x}{2 + \cos (2 π - 2 x)} dx = \int_{0}^{π} \frac{π - x}{2 + \cos (2 x)} dx

2 I = \int_{0}^{π} \frac{π}{2 + \cos (2 x)} dx = \int_{0}^{π} \frac{π}{1 + {2 \cos}^{2} x} dx

= \int_{0}^{π} \frac{π \sec^{2} x}{\sec^{2} x + 2} dx = \int_{0}^{π} \frac{π d (tanx)}{3 + \tan^{2} x}

= π {[\frac{1}{\sqrt{3}} Arctan (\frac{tanx}{\sqrt{3}})]}_{0}^{π} = \frac{π^{2}}{\sqrt{3}}

I = \frac{π^{2}}{2 \sqrt{3}} \approx 2.849109379

Commented by Lordose last updated on 25/Dec/20

\tan (π) = 0 bro

Commented by Ar Brandon last updated on 25/Dec/20

I realized sometime ago that when dealing with

Arctan (\frac{\tan (x)}{a}) {we}^{'} ve got to make use of the sign on zero .

That is \tan 0 = \frac{\sin 0}{\cos 0} = \frac{0}{+ 1} = + 0 \Rightarrow Arctan (\frac{\tan 0}{a}) = Arctan (+ 0) = 0

And same goes for \tan π

\tan π = \frac{\sin π}{\cos π} = \frac{0}{- 1} = - 0 then Arctan (\frac{\tan π}{a}) = Arctan (- 0) = π

{That}^{'} s why my answer corresponds with that obtained using

the scientific calculator . Perhabs someone may have a

better explanation for this result .

Commented by MJS_new last updated on 25/Dec/20

the function f (x) = \frac{π}{2 + \cos 2 x} obviously is

symmetric to x = \frac{π}{2} \Rightarrow

I = π \underset{0}{\int^{π / 2}} \frac{d x}{2 + \cos 2 x} =

[t = \tan x \to d x = \cos^{2} x d t]

= π \underset{0}{\int^{\infty}} \frac{d t}{t^{2} + 3} = \frac{π \sqrt{3}}{3} {[\arctan \frac{t}{\sqrt{3}}]}_{0}^{\infty} = \frac{π^{2} \sqrt{3}}{6} \approx 2.84911

Commented by MJS_new last updated on 25/Dec/20

\dots if the borders change from [a, b] to [c, c]

you always have to look for symmetry . that

is the point

Commented by Ar Brandon last updated on 26/Dec/20

Sounds better as an explanation.

Commented by Ar Brandon last updated on 26/Dec/20

Back from feast ? Haha !

Answered by Olaf last updated on 25/Dec/20

Ω = \int_{0}^{π} \frac{x}{2 + \cos (2 x)} d x (1)

Let u = π - x

2 Ω = \int_{π}^{0} \frac{π - u}{2 + \cos (2 u)} (- d u) = \int_{0}^{π} \frac{π - u}{2 + \cos (2 u)} d u (2)

(1) + (2) :

2 Ω = \int_{0}^{π} \frac{π}{2 + \cos (2 x)} d x

2 Ω = \int_{0}^{π} \frac{π}{2 + \frac{1 - \tan^{2} x}{1 + \tan^{2} x}} d x

2 Ω = \int_{0}^{π} \frac{π}{3 - \tan^{2} x} (1 + \tan^{2} x) d x

2 Ω = \int_{0}^{\frac{π}{2}} + \int_{\frac{π}{2}}^{π} = Ω_{1} + Ω_{2}

Let v = \tan x

Ω_{1} = π \int_{0}^{\infty} \frac{d v}{3 - v^{2}}

Ω_{1} = \frac{π}{2 \sqrt{3}} \int_{0}^{\infty} [\frac{1}{\sqrt{3} + v} + \frac{1}{\sqrt{3} - v}] d v

Ω_{1} = \frac{π}{2 \sqrt{3}} \int_{0}^{\infty} {[\ln ∣ \frac{\sqrt{3} + v}{\sqrt{3} - v} ∣]}_{0}^{\infty} = 0

\dots and Ω_{2} = 0

\Rightarrow Ω = 0

Commented by MJS_new last updated on 25/Dec/20

sorry but {you}^{'} re wrong

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