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

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$${\int }_0^{\mathbf{x}}\frac{{\mathrm{t}}^2}{{(\mathrm{t}\cdot \sinh \mathrm{t}-\cosh \mathrm{t})}^2}\mathrm{dt}=0$$

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

$$I=\int \frac{t^2}{{(t\sinh t-\cosh t)}^2}dt=\int \frac{t\cosh t}{{(t\sinh t-\cosh t)}^2}\cdot \frac{t}{\cosh t}dt$$$$\{\begin{array}{l}\mathrm{u}\left(t\right)=\frac{t}{\cosh t}\\ {\mathrm{v}}^{\prime }\left(t\right)=\frac{t\cosh t}{{(t\sinh t-\cosh t)}^2}\end{array}\Rightarrow \{\begin{array}{l}{\mathrm{u}}^{\prime }\left(t\right)=\frac{\cosh t-t\sinh t}{{\cosh }^{2}t}\\ \mathrm{v}\left(t\right)=-\frac{1}{t\sinh t-\cosh t}\end{array}$$$$I=-\frac{t}{\cosh t(t\sinh t-\cosh t)}-\int \frac{1}{{\cosh }^{2}t}dt$$$$=-\frac{t}{\cosh t(t\sinh t-\cosh t)}-\tanh t+C$$$${\int }_0^x\frac{t^2}{{(t\sinh t-\cosh t)}^2}dt=0$$$$\iff \frac{x}{\cosh x(\cosh x-x\sinh x)}-\tanh x=0$$$$\iff x-\sinh x(\cosh x-x\sinh x)=0$$$$\iff x+x{\sinh }^{2}x-\sinh x\cosh x=0$$$$\iff x(1+{\sinh }^{2}x)-\sinh x\cosh x=0$$$$\iff x{\cosh }^{2}x-\sinh x\cosh x=0$$$$\iff \cosh x(x\cosh x-\sinh x)=0,\cosh x⩾1\mathrm{\forall }x\in \mathbb{R}$$$$\iff x\cosh x-\sinh x=0\Rightarrow x=\tanh x$$

Commented by Shrinava last updated on 09/Jul/22

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

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