91Ó°ÊÓ

First, let's simplify the integrand. We know that \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\). So, $$\int_{1}^{\sqrt{3}} \frac{\operatorname{sech}(\ln x)}{x} d x = \int_{1}^{\sqrt{3}} \frac{1}{x \cosh(\ln x)} d x$$ Now, we also know that \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). Therefore, we can write the integral as $$\int_{1}^{\sqrt{3}} \frac{1}{x \left(\frac{e^{\ln x} + e^{-\ln x}}{2}\right)} d x = \int_{1}^{\sqrt{3}} \frac{2}{x (x + \frac{1}{x})} d x$$ Simplifying further, we get $$\int_{1}^{\sqrt{3}} \frac{2}{x^2 + 1} d x$$ To evaluate this integral, we'll use the arctangent integration formula: $$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \frac{x}{a} + C$$, with \(a = 1\) in this case. Therefore, the integral becomes $$\int_{1}^{\sqrt{3}}\frac{2}{x^2 +1}d x = 2\arctan x\bigg|_{1}^{\sqrt{3}}$$ Now, let's plug in our bounds: $$2(\arctan(\sqrt{3}) - \arctan(1)) = 2\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = 2\left(\frac{\pi}{12}\right) = \frac{\pi}{6}$$

In method b, we make the substitution \(u = \ln x\). The differential is \(du = \frac{1}{x} dx\). When \(x = 1\), \(u = \ln 1 = 0\). When \(x = \sqrt{3}\), \(u = \ln \sqrt{3}\). So, the integral becomes $$\int_{\ln1}^{\ln\sqrt{3}} \operatorname{sech}(u) d u$$ We know that the derivative of hyperbolic sine is the hyperbolic cosine, so we have $$\int_{\ln1}^{\ln\sqrt{3}} \operatorname{sech}(u) d u = 2 \tanh^{-1} (\tanh (u))\bigg|_{\ln1}^{\ln\sqrt{3}}$$ Elaborating further, $$ 2(\tanh^{-1} (\tanh (\ln\sqrt{3})) - \tanh^{-1} (\tanh (\ln1))) = 2(\tanh^{-1} (\frac{\sqrt{3}-1}{\sqrt{3}+1})-\tanh^{-1}(0)) = 2(\tanh^{-1}(\tanh(\frac{\pi}{6}))-0) = \frac{\pi}{6} $$ Both methods give the same result: \(\frac{\pi}{6}\).