∫_(−(π/2)) ^(π/2) ((sin 2x)/(1+2^x ))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 87977 by Ar Brandon last updated on 07/Apr/20

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{s i n 2 x}{1 + 2^{x}} d x$
Commented by abdomathmax last updated on 08/Apr/20
$at form of serie ∫_(−(π/2)) ^(π/2) ((sin(2x))/(1+2^x ))dx =∫_(−(π/2)) ^(π/2) ((2^(−x) sin(2x))/(1+2^(−x) ))dx =∫_(−(π/2)) ^(π/2) 2^(−x) sin(2x)Σ_(n=0) ^∞ (−1)^n 2^(−nx) dx =Σ_(n=0) ^∞ (−1)^n ∫_(−(π/2)) ^(π/2) 2^(−(n+1)x) sin(2x)dx but ∫_(−(π/2)) ^(π/2) 2^(−(n+1)x) sin(2x)dx =Im(∫_(−(π/2)) ^(π/2) 2^(−(n+1)x) e^(2ix) dx) ∫_(−(π/2)) ^(π/2) e^(−(n+1)xln2 +2ix) dx =∫_(−(π/2)) ^(π/2) e^((2i−(n+1)ln2)x) dx =[(1/(2i−(n+1)ln2)) e^((2i−(n+1)ln2)x) ]_(−(π/2)) ^(π/2) =(1/(2i−(n+1)ln2)){ −e^(−(n+1)ln2(π/2)) −e^((n+1)ln2(π/2)) } =(((n+1)ln2−2i)/((n+1)^2 ln^2 2+4)){ e^((π/2)(n+1)ln2) +e^(−(π/2)(n+1)ln2) } ⇒ Im(...) =−(2/((n+1)^2 ln^2 2+4)) ×2 ch((π/2)(n+1)ln2) ⇒ ⇒ I =4Σ_(n=0) ^∞ (((−1)^(n+1) )/((n+1)^2 ln^2 2 +4))×ch((π/2)(n+1)ln2)$
$a t f o r m o f s e r i e$ $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{s i n (2 x)}{1 + 2^{x}} d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{2^{- x} s i n (2 x)}{1 + 2^{- x}} d x$ $= \int_{- \frac{π}{2}}^{\frac{π}{2}} 2^{- x} s i n (2 x) \sum_{n = 0}^{\infty} {(- 1)}^{n} 2^{- n x} d x$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{- \frac{π}{2}}^{\frac{π}{2}} 2^{- (n + 1) x} s i n (2 x) d x$ $b u t \int_{- \frac{π}{2}}^{\frac{π}{2}} 2^{- (n + 1) x} s i n (2 x) d x$ $= I m (\int_{- \frac{π}{2}}^{\frac{π}{2}} 2^{- (n + 1) x} e^{2 i x} d x)$ $\int_{- \frac{π}{2}}^{\frac{π}{2}} e^{- (n + 1) x l n 2 + 2 i x} d x$ $= \int_{- \frac{π}{2}}^{\frac{π}{2}} e^{(2 i - (n + 1) l n 2) x} d x$ $= {[\frac{1}{2 i - (n + 1) l n 2} e^{(2 i - (n + 1) l n 2) x}]}_{- \frac{π}{2}}^{\frac{π}{2}}$ $= \frac{1}{2 i - (n + 1) l n 2} {- e^{- (n + 1) l n 2 \frac{π}{2}} - e^{(n + 1) l n 2 \frac{π}{2}}}$ $= \frac{(n + 1) l n 2 - 2 i}{{(n + 1)}^{2} {l n}^{2} 2 + 4} {e^{\frac{π}{2} (n + 1) l n 2} + e^{- \frac{π}{2} (n + 1) l n 2}} \Rightarrow$ $I m (. . .) = - \frac{2}{{(n + 1)}^{2} {l n}^{2} 2 + 4} \times 2 c h (\frac{π}{2} (n + 1) l n 2) \Rightarrow$ $\Rightarrow$ $I = 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{{(n + 1)}^{2} {l n}^{2} 2 + 4} \times c h (\frac{π}{2} (n + 1) l n 2)$
Commented by Ar Brandon last updated on 08/Apr/20

$A m a z i n g!!!!!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum