Question Number 37338 by math khazana by abdo last updated on 12/Jun/18
$${calculate}\:\:{B}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{sh}^{{n}} {xdx}\:. \\ $$
Commented by prof Abdo imad last updated on 15/Jun/18
![B_n = ∫_0 ^1 { ((e^x −e^(−x) )/2)}^n dx = (1/2^n ) ∫_0 ^1 {Σ_(k=0) ^n C_n ^k (e^x )^k (−e^(−x) )^(n−k) }dx =(1/2^n ) ∫_0 ^1 Σ_(k=0) ^n C_n ^k e^(kx) (−1)^(n−k) e^(−(n−k)) dx = (1/2^n ) Σ_(k=0) ^n (−1)^(n−k) C_n ^k ∫_0 ^1 e^((2k−n)x) dx =(1/2^n ) Σ_(k=0) ^n (−1)^(n−k) C_n ^k [ (1/(2k−n)) e^((2k−n)x) ]_0 ^1 B_n = (1/2^n ) Σ_(k=0) ^n (−1)^(n−k) (C_n ^k /(2k−n)){ e^(2k−n) −1} .](https://www.tinkutara.com/question/Q37617.png)
$${B}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\left\{\:\frac{{e}^{{x}} \:−{e}^{−{x}} }{\mathrm{2}}\right\}^{{n}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left\{\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({e}^{{x}} \right)^{{k}} \:\left(−{e}^{−{x}} \right)^{{n}−{k}} \right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{e}^{{kx}} \left(−\mathrm{1}\right)^{{n}−{k}} \:{e}^{−\left({n}−{k}\right)} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{n}−{k}} \:{C}_{{n}} ^{{k}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{\left(\mathrm{2}{k}−{n}\right){x}} \:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{n}−{k}} \:{C}_{{n}} ^{{k}} \:\:\left[\:\frac{\mathrm{1}}{\mathrm{2}{k}−{n}}\:{e}^{\left(\mathrm{2}{k}−{n}\right){x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${B}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{n}−{k}} \:\:\frac{{C}_{{n}} ^{{k}} }{\mathrm{2}{k}−{n}}\left\{\:{e}^{\mathrm{2}{k}−{n}} \:−\mathrm{1}\right\}\:. \\ $$