calculate-I-n-0-pi-dx-1-cos-2-nx-

Question Number 36936 by maxmathsup by imad last updated on 07/Jun/18

c a l c u l a t e I_{n} = \int_{0}^{π} \frac{d x}{1 + {c o s}^{2} (n x)}

Commented by math khazana by abdo last updated on 02/Aug/18

w e h a v e I_{n} = \int_{0}^{π} \frac{d x}{1 + \frac{1 + c o s (2 n x)}{2}}

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

= \frac{1}{n} \int_{0}^{2 n π} \frac{d t}{3 + c o s t} \Rightarrow {n I}_{n} = \sum_{k = 0}^{n - 1} \int_{2 k π}^{2 (k + 1) π} \frac{d t}{3 + c o s t}

=_{t = 2 k π + x} \sum_{k = 0}^{n - 1} \int_{0}^{2 π} \frac{d x}{3 + c o s (x + 2 k π)}

= n \int_{0}^{2 π} \frac{d x}{3 + c o s x} \Rightarrow I_{n} = \int_{0}^{2 π} \frac{d x}{3 + c o s x} \forall n ⩾ 1 a n d

I_{o} = π c h a n g e m e n t e^{i x} = z g i v e

I_{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} > 1 \Rightarrow

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

φ (z) = \frac{- 2 i}{(z - z_{1}) (z - z_{2})} \Rightarrow 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

I_{n} = \frac{π}{\sqrt{2}} w i t h n ⩾ 1 .

Commented by math khazana by abdo last updated on 02/Aug/18

I_{0} = \frac{π}{2}