calculate ∫_(−1) ^1 (x^(2n) /(1+2^(sinx) ))dx with n integr.


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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
	$∫_(−1) ^1 (x^(2n) /(1+2^(sin(x)) ))dx=∫_(−1) ^1 (((−x)^(2n) )/(1+2^(sin(−x)) ))dx=∫_(−1) ^1 (x^(2n) /(1+2^(−sin(x)) ))dx=∫_(−1) ^1 ((2^(sin(x)) x^(2n) )/(1+2^(son(x)) )) ==>2∫_(−1) ^1 (x^(2n) /(1+2^(sin(x)) )) dx=∫_(−1) ^1 (x^(2n) /(1+2^(sin(x)) ))dx+∫_(−1) ^1 ((x^(2n) 2^(sin(x{) )/(1+2^(sin(x)) ))dx=∫_(−1) ^1 ((x^(2n) +x^(2n) 2^(sin(x)) )/(1+2^(sin(x)) ))=∫_(−1) ^1 x^(2n) dx=(2/(2n+1)) =>∫_(−1) ^1 (x^(2n) /(1+2^(sin(x)) ))dx=(1/(2n+1))$
	$\int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n (x)}} d x = \int_{- 1}^{1} \frac{{(- x)}^{2 n}}{1 + 2^{s i n (- x)}} d x = \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{- s i n (x)}} d x = \int_{- 1}^{1} \frac{2^{s i n (x)} x^{2 n}}{1 + 2^{s o n (x)}}$ $==> 2 \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n (x)}} d x = \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n (x)}} d x + \int_{- 1}^{1} \frac{x^{2 n} 2^{s i n (x {}}{1 + 2^{s i n (x)}} d x = \int_{- 1}^{1} \frac{x^{2 n} + x^{2 n} 2^{s i n (x)}}{1 + 2^{s i n (x)}} = \int_{- 1}^{1} x^{2 n} d x = \frac{2}{2 n + 1}$ $=> \int_{- 1}^{1} \frac{x^{2 n}}{1 + 2^{s i n (x)}} d x = \frac{1}{2 n + 1}$
	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

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