From the previous figure we have \[\tan\theta=\frac{\text{d}y}{\text{d}x}\] because \(\tan\theta\) is the slope of the tangent to curve at the point \(M\left(x,y\right)\). \[\] The condition for the static equilibrium indicates that \[\sum F_{x}=0\Rightarrow T_{0}=T\cos\theta\] We have an uniform distribution of weights therefore the global weight is equal to \(w_{0}l\) \[\sum F_{y}=0\Rightarrow w_{0}l=T\sin\theta\] From the two previous results calculated using the static equilibrium condition, we get \[\frac{w_{0}l}{T_{0}}=\tan\theta=\frac{\text{d}y}{\text{d}x}\] Let \(a=\frac{w_{0}}{T_{0}}\), \[\Rightarrow\frac{\text{d}y}{\text{d}x}=al\] Recall: The length of a curve arc can be found by \[\color{orange} {l=\int_{}^{} \sqrt{1+\left(\frac{\text{d}y}{\text{d}x}\right)^{2}}\text{d}x}\] \[\Rightarrow\frac{\text{d}l}{\text{d}x}= \sqrt{1+\left(\frac{\text{d}y}{\text{d}x}\right)^{2}}\] We have, \[\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right)= \frac{\text{d}}{\text{d}x}\left(al\right)\] \[\frac{d^{2}y}{dx^{2}}=a\frac{\text{d}l}{\text{d}x}=a\sqrt{1+\left(\frac{\text{d}y}{\text{d}x}\right)^{2}}\] Let \[P=\frac{\text{d}y}{\text{d}x}\] \[\Rightarrow \frac{\text{d}P}{\text{d}x}=a\sqrt{1+P^{2}}\] \[\Rightarrow \frac{1}{\sqrt{1+P^{2}}}\text{d}P=a\text{d}x\] Thus, \[\int_{}^{} \frac{1}{\sqrt{1+P^{2}}}\text{d}P=\int_{}^{}a\text{d}x=ax+c_{2}\] Now let’s calculate \(\int_{}^{} \frac{1}{\sqrt{1+P^{2}}}\text{d}P\), \[J=\int_{}^{} \frac{1}{\sqrt{1+P^{2}}}\text{d}P\] Let \(P=\tan\alpha\Rightarrow\text{d}P=\left(\sec\alpha\right)^{2}\text{d}\alpha\) \[J=\int_{}^{} \frac{1}{\sqrt{1+\left(\tan\alpha\right)^{2}}}\left(\sec\alpha\right)^{2}\text{d}\alpha\] We have \(1+\left(\tan\alpha\right)^{2}=\left(\sec\alpha\right)^{2}\) \[J=\log\mid\sec\alpha+\tan\alpha\mid\] We have \(\tan\alpha=P\) and \(\sec\alpha=\sqrt{1+\left(\tan\alpha\right)^{2}}=\sqrt{1+P^{2}}\) \[\Rightarrow J=\log\mid P+\sqrt{1+P^{2}}\mid+c_{1}\] \(P=\frac{\text{d}y}{\text{d}x}\geq0\) because the curve is increasing when \(x\geq0\) (the graph is symmetric about \(y-axis\) it is enough to deal with \(x\geq0\)) and we have \(\sqrt{1+P^{2}}>0\) therefore, \[J=\log \left(P+\sqrt{1+P^{2}}\right)+c_{1}\] Now, let’s get back to the main problem \[\int_{}^{} \frac{1}{\sqrt{1+P^{2}}}\text{d}P=\int_{}^{}a\text{d}x=ax+c_{2}\] \[\Rightarrow\log \left(P+\sqrt{1+P^{2}}\right)+c_{1}=ax+c_{2}\] \[\Rightarrow\log \left(P+\sqrt{1+P^{2}}\right)=ax+c_{3}\] For \(x=0\Rightarrow\frac{\text{d}y}{\text{d}x}=P=0\Rightarrow\log \left(0+\sqrt{1+0^{2}}\right)=a0+c_{3}\Rightarrow c_{3}=0\) Thus, \[\log \left(P+\sqrt{1+P^{2}}\right)=ax\Rightarrow P+\sqrt{1+P^{2}}=e^{ax}\] Using conjugate, we get \[e^{ax}\sqrt{1+P^{2}}-e^{ax}P=1\] \[\Rightarrow\left(e^{ax}\sqrt{1+P^{2}}\right)^{2}=\left(1+e^{ax}P\right)^{2}\] \[\Rightarrow1+2e^{ax}P=e^{2ax}\] \[\Rightarrow P=\frac{e^{2ax}-1}{2e^{ax}}\] Therefore \[\Rightarrow P=\frac{\text{d}y}{\text{d}x}=\frac{e^{ax}-e^{-ax}}{2}\] \[\Rightarrow \int_{}^{} \text{d}y=\int_{}^{} \frac{e^{ax}-e^{-ax}}{2}\text{d}x\] Finally \[\color{green} {\large y=\frac{\cosh\left(ax\right)}{a}+c}\]