Question-138580

Question Number 138580 by Bekzod Jumayev last updated on 15/Apr/21

Commented by Bekzod Jumayev last updated on 15/Apr/21

P l e a s e h e l p ?

Answered by Ar Brandon last updated on 15/Apr/21

\int_{a}^{b} f (x) dx = \int_{a}^{b} f (a + b - x) dx

\int_{- π}^{π} \frac{\sin (2021 x)}{(1 + 3^{x}) sinx} dx = \int_{- π}^{π} \frac{\sin (2021 x)}{(1 + 3^{- x}) sinx} dx = \int_{- π}^{π} \frac{3^{x} \sin (2021 x)}{(1 + 3^{x}) sinx} dx

= \frac{1}{2} \int_{- π}^{π} \frac{(1 + 3^{x}) \sin (2021 x)}{(1 + 3^{x}) sinx} dx = \frac{1}{2} \int_{- π}^{π} \frac{\sin (2021 x)}{sinx} dx

= \int_{0}^{π} \frac{\sin (2021 x)}{sinx} dx

Answered by phanphuoc last updated on 15/Apr/21

I = \int_{- π}^{π} 3^{x} s i n (2021 x) d x / (1 + 3^{x}) s i n x w i t h x = - u

2 I = \int_{- π}^{π} s i n (2021 x) d x / s i n x

Answered by mnjuly1970 last updated on 15/Apr/21

ϕ_{n} = \int_{0}^{π} \frac{s i n (n x)}{s i n (x)} \dots \dots . . (*)

ϕ_{n - 2} = \int_{0}^{π} \frac{s i n ((n - 2) x)}{s i n (x)} d x

ϕ_{n} - ϕ_{n - 2} = \int_{0}^{π} \frac{2 c o s (\frac{n x + n x - 2 x}{2}) s i n (\frac{2 x}{2})}{s i n (x)} d x

= \int_{0}^{π} \frac{c o s ((n - 1) x) . s i n (x)}{s i n (x)} d x

= \frac{1}{n - 1} \int_{0}^{π} c o s ((n - 1) x) d x = 0

ϕ_{n} = ϕ_{n - 2} \dots \dots ✓

(*) \dots \dots :: ϕ_{0} = 0, ϕ_{1} = π, ϕ_{2} = 0 (✓) . .

ϕ_{n} = {\begin{cases} 0 i f; n := e v e n \\ π i f; n := o d d \end{cases}

\dots \dots \dots . ∴ ϕ_{2021} = π \dots \dots \dots .