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Question Number 173250 by Shrinava last updated on 08/Jul/22

Solve for real numbers: ∫_0 ^( x) (t^2 /((t ∙ sinh t − cosh t)^2 )) dt = 0

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\int_{\mathrm{0}} ^{\:\boldsymbol{\mathrm{x}}} \:\frac{\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{t}\:\centerdot\:\mathrm{sinh}\:\mathrm{t}\:−\:\mathrm{cosh}\:\mathrm{t}\right)^{\mathrm{2}} }\:\mathrm{dt}\:=\:\mathrm{0} \\ $$

Answered by Ar Brandon last updated on 09/Jul/22

I=∫(t^2 /((tsinht−cosht)^2 ))dt=∫((tcosht)/((tsinht−cosht)^2 ))∙(t/(cosht))dt { ((u(t)=(t/(cosht)))),((v′(t)=((tcosht)/((tsinht−cosht)^2 )))) :} ⇒ { ((u′(t)=((cosht−tsinht)/(cosh^2 t)))),((v(t)=−(1/(tsinht−cosht)))) :} I=−(t/(cosht(tsinht−cosht)))−∫(1/(cosh^2 t))dt =−(t/(cosht(tsinht−cosht)))−tanht+C ∫_0 ^x (t^2 /((tsinht−cosht)^2 ))dt=0 ⇔(x/(coshx(coshx−xsinhx)))−tanhx=0 ⇔x−sinhx(coshx−xsinhx)=0 ⇔x+xsinh^2 x−sinhxcoshx=0 ⇔x(1+sinh^2 x)−sinhxcoshx=0 ⇔xcosh^2 x−sinhxcoshx=0 ⇔coshx(xcoshx−sinhx)=0, coshx ≥ 1 ∀x∈R ⇔xcoshx−sinhx=0⇒x=tanhx

$${I}=\int\frac{{t}^{\mathrm{2}} }{\left({t}\mathrm{sinh}{t}−\mathrm{cosh}{t}\right)^{\mathrm{2}} }{dt}=\int\frac{{t}\mathrm{cosh}{t}}{\left({t}\mathrm{sinh}{t}−\mathrm{cosh}{t}\right)^{\mathrm{2}} }\centerdot\frac{{t}}{\mathrm{cosh}{t}}{dt} \\ $$$$\:\:\begin{cases}{\mathrm{u}\left({t}\right)=\frac{{t}}{\mathrm{cosh}{t}}}\\{\mathrm{v}'\left({t}\right)=\frac{{t}\mathrm{cosh}{t}}{\left({t}\mathrm{sinh}{t}−\mathrm{cosh}{t}\right)^{\mathrm{2}} }}\end{cases}\:\Rightarrow\begin{cases}{\mathrm{u}'\left({t}\right)=\frac{\mathrm{cosh}{t}−{t}\mathrm{sinh}{t}}{\mathrm{cosh}^{\mathrm{2}} {t}}}\\{\mathrm{v}\left({t}\right)=−\frac{\mathrm{1}}{{t}\mathrm{sinh}{t}−\mathrm{cosh}{t}}}\end{cases} \\ $$$${I}=−\frac{{t}}{\mathrm{cosh}{t}\left({t}\mathrm{sinh}{t}−\mathrm{cosh}{t}\right)}−\int\frac{\mathrm{1}}{\mathrm{cosh}^{\mathrm{2}} {t}}{dt} \\ $$$$\:\:\:=−\frac{{t}}{\mathrm{cosh}{t}\left({t}\mathrm{sinh}{t}−\mathrm{cosh}{t}\right)}−\mathrm{tanh}{t}+{C} \\ $$$$\int_{\mathrm{0}} ^{{x}} \frac{{t}^{\mathrm{2}} }{\left({t}\mathrm{sinh}{t}−\mathrm{cosh}{t}\right)^{\mathrm{2}} }{dt}=\mathrm{0} \\ $$$$\Leftrightarrow\frac{{x}}{\mathrm{cosh}{x}\left(\mathrm{cosh}{x}−{x}\mathrm{sinh}{x}\right)}−\mathrm{tanh}{x}=\mathrm{0} \\ $$$$\Leftrightarrow{x}−\mathrm{sinh}{x}\left(\mathrm{cosh}{x}−{x}\mathrm{sinh}{x}\right)=\mathrm{0} \\ $$$$\Leftrightarrow{x}+{x}\mathrm{sinh}^{\mathrm{2}} {x}−\mathrm{sinh}{x}\mathrm{cosh}{x}=\mathrm{0} \\ $$$$\Leftrightarrow{x}\left(\mathrm{1}+\mathrm{sinh}^{\mathrm{2}} {x}\right)−\mathrm{sinh}{x}\mathrm{cosh}{x}=\mathrm{0} \\ $$$$\Leftrightarrow{x}\mathrm{cosh}^{\mathrm{2}} {x}−\mathrm{sinh}{x}\mathrm{cosh}{x}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{cosh}{x}\left({x}\mathrm{cosh}{x}−\mathrm{sinh}{x}\right)=\mathrm{0},\:\mathrm{cosh}{x}\:\geqslant\:\mathrm{1}\:\forall{x}\in\mathbb{R} \\ $$$$\Leftrightarrow{x}\mathrm{cosh}{x}−\mathrm{sinh}{x}=\mathrm{0}\Rightarrow{x}=\mathrm{tanh}{x} \\ $$

Commented by Shrinava last updated on 09/Jul/22

thank you professor, answer.?

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor},\:\mathrm{answer}.? \\ $$

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