1-calculate-U-n-0-pi-dx-1-cos-2-nx-with-n-from-N-2-f-continue-from-0-pi-to-R-find-lim-n-0-pi-f-x-1-cos-2-nx-dx-

Question Number 50412 by Abdo msup. last updated on 16/Dec/18

1) c a l c u l a t e U_{n} = \int_{0}^{π} \frac{d x}{1 + {c o s}^{2} (n x)} w i t h n f r o m N

2) f c o n t i n u e f r o m [0, π] t o R f i n d

{l i m}_{n \to + \infty} \int_{0}^{π} \frac{f (x)}{1 + {c o s}^{2} (n x)} d x

Commented by Abdo msup. last updated on 25/Dec/18

1) c h a n g e m e n t n x = t g i v e

U_{n} = \frac{1}{n} \int_{0}^{n π} \frac{1}{1 + {c o s}^{2} t} d t \Rightarrow {n U}_{n} = \sum_{k = 0}^{n - 1} \int_{k π}^{(k + 1) π} \frac{d t}{1 + {c o s}^{2} t}

b u t \int_{k π}^{(k + 1) π} \frac{d t}{1 + {c o s}^{2} t} =_{t = k π + x} \int_{0}^{π} \frac{d x}{1 + {c o s}^{2} x}

= \int_{0}^{π} \frac{d x}{1 + \frac{1 + c o s (2 x)}{2}} = \int_{0}^{π} \frac{2 d x}{3 + c o s (2 x)}

=_{2 x = u} \int_{0}^{2 π} \frac{d u}{3 + c o s u} d u \Rightarrow {n U}_{n} = n \int_{0}^{2 π} \frac{d u}{3 + c o s u}

\Rightarrow U_{n} = \int_{0}^{2 π} \frac{d u}{3 + c o s (u)} c h a n g e m e n t e^{i u} = z g i v e

U_{n} = \int_{∣ z ∣= 1} \frac{1}{3 + \frac{z + z^{- 1}}{2}} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{2 d z}{i z (6 + z + z^{- 1})} = \int_{∣ z ∣= 1} \frac{- 2 i d z}{6 z + z^{2} + 1}

l e t φ (z) = \frac{- 2 i}{z^{2} + 6 z + 1} p o l e s o f φ ?

Δ^{^{'}} = 3^{2} - 1 = 8 \Rightarrow z_{1} = - 3 + 2 \sqrt{2} a n d z_{2} = - 3 - 2 \sqrt{2}

∣ z_{1} ∣ - 1 = 3 - 2 \sqrt{2} - 1 = 2 - 2 \sqrt{2} < 0 \Rightarrow∣ z_{1} ∣< 1

∣ z_{2} ∣ - 1 = 3 + 2 \sqrt{2} - 1 = 2 + 2 \sqrt{2} (z_{2} i s o u t o f c i r c l e)

\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1})

R e s (φ, z_{1}) = \frac{- 2 i}{z_{1} - z_{2}} = \frac{- 2 i}{4 \sqrt{2}} = \frac{- i}{2 \sqrt{2}} \Rightarrow

\int_{∣ z ∣= 1} φ (z) d z = 2 i π (\frac{- i}{2 \sqrt{2}}) = \frac{π}{\sqrt{2}}

\Rightarrow \forall n \in N U_{n} = \frac{π}{\sqrt{2}} .