0-2pi-z-sin-z-1-cos-2-z-dz-z-2-1-z-2-1-dz-z-2-sin-z-z-2-1-dz-

Question Number 213518 by issac last updated on 07/Nov/24

\int_{0}^{2 π} \frac{z \cdot \sin (z)}{1 + \cos^{2} (z)} d z

\int_{∣ z ∣= 2} \frac{1}{z^{2} + 1} d z

\int_{∣ z ∣= 2} \frac{\sin (z)}{z^{2} + 1} d z

Answered by Berbere last updated on 07/Nov/24

\int_{0}^{π} \frac{z s i n (z)}{1 + {c o s}^{2} (z)} + \int_{0}^{π} \frac{s i n (π + z) (π + z) d z}{1 + {c o z}^{2} (z)}

{= - π \int_{0}^{π} \frac{s i n (z)}{1 + {c o s}^{2} (z)} d z = π \tan^{- 1} (c o s (z))]}_{0}^{π}

= - \frac{π^{2}}{2}

2 . . = 2 i π R e s (\frac{1}{z^{2} + 1}, i; - i} = 2 i π . (\frac{1}{2 i} - \frac{1}{2 i}) = 0

. . 3

= \int \frac{s i n (z)}{1 + z^{2}} d z = 2 i π R e s (\frac{s i n (z)}{1 + z^{2}}, z = i, - i)

= 2 i π [\frac{s i n (i)}{2 i} + \frac{s i n (- i)}{- 2 i})

= 2 π s i n (i) = - 2 π s h (1)

Commented by York12 last updated on 08/Nov/24

{What}^{'} s your instgram or facebook account

, please sir

Commented by issac last updated on 08/Nov/24

thx \dots

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