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calculate-0-lln-3-sh-2-x-dx-ch-3-x-




Question Number 50418 by Abdo msup. last updated on 16/Dec/18
calculate ∫_0 ^(lln(3))   ((sh^2 (x)dx)/(ch^3 (x)))
$${calculate}\:\int_{\mathrm{0}} ^{{lln}\left(\mathrm{3}\right)} \:\:\frac{{sh}^{\mathrm{2}} \left({x}\right){dx}}{{ch}^{\mathrm{3}} \left({x}\right)} \\ $$
Commented by Abdo msup. last updated on 24/Dec/18
let A =∫_0 ^(ln(3))   ((sh^2 (x))/(ch^3 (x)))dx  we have  A = ∫_0 ^(ln(3))  ((ch^2 x−1)/(ch^3 x))dx =∫_0 ^(ln(3))  (dx/(ch(x))) −∫_0 ^(ln(3))  (dx/(ch^3 x))  ∫_0 ^(ln(3))   (dx/(ch(x))) =∫_0 ^(ln(3))    ((2dx)/(e^x  +e^(−x) )) =_(e^x =t)   ∫_1 ^3    (2/(t+t^(−1) )) (dt/t)  = ∫_1 ^3    ((2dt)/(t^2  +1)) =[2arctan(t)]_1 ^3   =2( arctan(3)−(π/4))=2arctan(3)−(π/2)  ∫_0 ^(ln(3))    (dx/(ch^3 x)) =8 ∫_0 ^(ln(3))   (dx/((e^x  +e^(−x) )^3 ))  =8 ∫_0 ^(ln(3))    (dx/(e^(3x)  +3 e^(2x)  e^(−x)  +3 e^x  e^(−2x)  +e^(−3x) ))  =8 ∫_0 ^(ln(3)   )      (dx/(e^(3x)   +e^(−3x)  +3 e^(x )  +3 e^(−x) ))  =_(e^x =t)  8 ∫_1 ^3         (dt/(t(t^3  +t^(−3)  +3t +3t^(−1) )))  =8 ∫_1 ^3    (dt/(t^4  +t^(−2)  +3t^2  +3))  =8 ∫_1 ^3     ((t^2 dt)/(t^(6 )  +1 +3t^4  +3t^2 ))  let decompose  F(t) = (t^2 /(t^6  +3t^4  +3t^2  +1))  ....be continued....
$${let}\:{A}\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)} \:\:\frac{{sh}^{\mathrm{2}} \left({x}\right)}{{ch}^{\mathrm{3}} \left({x}\right)}{dx}\:\:{we}\:{have} \\ $$$${A}\:=\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)} \:\frac{{ch}^{\mathrm{2}} {x}−\mathrm{1}}{{ch}^{\mathrm{3}} {x}}{dx}\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)} \:\frac{{dx}}{{ch}\left({x}\right)}\:−\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)} \:\frac{{dx}}{{ch}^{\mathrm{3}} {x}} \\ $$$$\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)} \:\:\frac{{dx}}{{ch}\left({x}\right)}\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)} \:\:\:\frac{\mathrm{2}{dx}}{{e}^{{x}} \:+{e}^{−{x}} }\:=_{{e}^{{x}} ={t}} \:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\frac{\mathrm{2}}{{t}+{t}^{−\mathrm{1}} }\:\frac{{dt}}{{t}} \\ $$$$=\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\left[\mathrm{2}{arctan}\left({t}\right)\right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$=\mathrm{2}\left(\:{arctan}\left(\mathrm{3}\right)−\frac{\pi}{\mathrm{4}}\right)=\mathrm{2}{arctan}\left(\mathrm{3}\right)−\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)} \:\:\:\frac{{dx}}{{ch}^{\mathrm{3}} {x}}\:=\mathrm{8}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)} \:\:\frac{{dx}}{\left({e}^{{x}} \:+{e}^{−{x}} \right)^{\mathrm{3}} } \\ $$$$=\mathrm{8}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)} \:\:\:\frac{{dx}}{{e}^{\mathrm{3}{x}} \:+\mathrm{3}\:{e}^{\mathrm{2}{x}} \:{e}^{−{x}} \:+\mathrm{3}\:{e}^{{x}} \:{e}^{−\mathrm{2}{x}} \:+{e}^{−\mathrm{3}{x}} } \\ $$$$=\mathrm{8}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{3}\right)\:\:\:} \:\:\:\:\:\frac{{dx}}{{e}^{\mathrm{3}{x}} \:\:+{e}^{−\mathrm{3}{x}} \:+\mathrm{3}\:{e}^{{x}\:} \:+\mathrm{3}\:{e}^{−{x}} } \\ $$$$=_{{e}^{{x}} ={t}} \:\mathrm{8}\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\:\:\:\:\:\frac{{dt}}{{t}\left({t}^{\mathrm{3}} \:+{t}^{−\mathrm{3}} \:+\mathrm{3}{t}\:+\mathrm{3}{t}^{−\mathrm{1}} \right)} \\ $$$$=\mathrm{8}\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\frac{{dt}}{{t}^{\mathrm{4}} \:+{t}^{−\mathrm{2}} \:+\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{3}} \\ $$$$=\mathrm{8}\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\:\frac{{t}^{\mathrm{2}} {dt}}{{t}^{\mathrm{6}\:} \:+\mathrm{1}\:+\mathrm{3}{t}^{\mathrm{4}} \:+\mathrm{3}{t}^{\mathrm{2}} } \\ $$$${let}\:{decompose}\:\:{F}\left({t}\right)\:=\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{6}} \:+\mathrm{3}{t}^{\mathrm{4}} \:+\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$….{be}\:{continued}…. \\ $$$$ \\ $$

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