calculate-1-1-x-2n-1-2-sinx-dx-with-n-integr-

Question Number 67038 by mathmax by abdo last updated on 22/Aug/19

c a l c u l a t e \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n x}} d x w i t h n i n t e g r .

Commented by mathmax by abdo last updated on 22/Aug/19

l e t A_{n} = \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n x}} d x \Rightarrow A_{n} = \int_{- 1}^{0} \frac{x^{2 n}}{1 + 2^{s i n x}} d x + \int_{0}^{1} \frac{x^{2 n}}{1 + 2^{s i n x}} d x b u t

\int_{- 1}^{0} \frac{x^{2 n}}{1 + 2^{s i n x}} d x =_{x = - t} \int_{0}^{1} \frac{t^{2 n}}{1 + 2^{- s i n t}} d t = \int_{0}^{1} \frac{2^{s i n t} t^{2 n}}{2^{s i n t} + 1} d t \Rightarrow

A_{n} = \int_{0}^{1} (\frac{2^{s i n t} t^{2 n}}{1 + 2^{s i n t}} + \frac{t^{2 n}}{1 + 2^{s i n t}}) d t = \int_{0}^{1} t^{2 n} (\frac{1 + 2^{s i n t}}{1 + 2^{s i n t}}) d t = \int_{0}^{1} t^{2 n} d t

{[\frac{t^{2 n + 1}}{2 n + 1}]}_{0}^{1} = \frac{1}{2 n + 1} \Rightarrow ★ A_{n} = \frac{1}{2 n + 1} ★

Answered by mind is power last updated on 22/Aug/19

\int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n (x)}} d x = \int_{- 1}^{1} \frac{{(- x)}^{2 n}}{1 + 2^{s i n (- x)}} d x = \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{- s i n (x)}} d x = \int_{- 1}^{1} \frac{2^{s i n (x)} x^{2 n}}{1 + 2^{s o n (x)}}

==> 2 \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n (x)}} d x = \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n (x)}} d x + \int_{- 1}^{1} \frac{x^{2 n} 2^{s i n (x {}}{1 + 2^{s i n (x)}} d x = \int_{- 1}^{1} \frac{x^{2 n} + x^{2 n} 2^{s i n (x)}}{1 + 2^{s i n (x)}} = \int_{- 1}^{1} x^{2 n} d x = \frac{2}{2 n + 1}

=> \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n (x)}} d x = \frac{1}{2 n + 1}

Commented by mathmax by abdo last updated on 22/Aug/19

t h a n k y o u s i r .

Commented by mind is power last updated on 23/Aug/19

y^{'} r e w e l c o m