Solve-for-real-numbers-0-x-t-2-t-sinh-t-cosh-t-2-dt-0-

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

Solve for real numbers :

\int_{0}^{x} \frac{t^{2}}{{(t \cdot \sinh t - \cosh t)}^{2}} dt = 0

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

I = \int \frac{t^{2}}{{(t \sinh t - \cosh t)}^{2}} d t = \int \frac{t \cosh t}{{(t \sinh t - \cosh t)}^{2}} \cdot \frac{t}{\cosh t} d t

{\begin{cases} u (t) = \frac{t}{\cosh t} \\ v^{'} (t) = \frac{t \cosh t}{{(t \sinh t - \cosh t)}^{2}} \end{cases} \Rightarrow {\begin{cases} u^{'} (t) = \frac{\cosh t - t \sinh t}{\cosh^{2} t} \\ v (t) = - \frac{1}{t \sinh t - \cosh t} \end{cases}

I = - \frac{t}{\cosh t (t \sinh t - \cosh t)} - \int \frac{1}{\cosh^{2} t} d t

= - \frac{t}{\cosh t (t \sinh t - \cosh t)} - \tanh t + C

\int_{0}^{x} \frac{t^{2}}{{(t \sinh t - \cosh t)}^{2}} d t = 0

\Leftrightarrow \frac{x}{\cosh x (\cosh x - x \sinh x)} - \tanh x = 0

\Leftrightarrow x - \sinh x (\cosh x - x \sinh x) = 0

\Leftrightarrow x + x \sinh^{2} x - \sinh x \cosh x = 0

\Leftrightarrow x (1 + \sinh^{2} x) - \sinh x \cosh x = 0

\Leftrightarrow x \cosh^{2} x - \sinh x \cosh x = 0

\Leftrightarrow \cosh x (x \cosh x - \sinh x) = 0, \cosh x ⩾ 1 \forall x \in R

\Leftrightarrow x \cosh x - \sinh x = 0 \Rightarrow x = \tanh x

Commented by Shrinava last updated on 09/Jul/22

thank you professor, answer . ?