Finley Mckinney

2022-06-19

Evaluate
${\int }_{\mathrm{ln}2}^{\mathrm{ln}4}\frac{{e}^{-x}}{\sqrt{1-{e}^{-2x}}}dx$

Nola Rivera

Expert

Observe that
${{\int }_{1/4}^{1/2}\frac{du}{\sqrt{1-{u}^{2}}}=\mathrm{arcsin}y|}_{1/4}^{1/2}=\mathrm{arcsin}\frac{1}{2}-\mathrm{arcsin}\frac{1}{4}$
But $\phantom{\rule{thickmathspace}{0ex}}\mathrm{arcsin}x=\mathrm{arctan}\frac{x}{\sqrt{1-{x}^{2}}}\phantom{\rule{thickmathspace}{0ex}}$ , so the above result equals
$\mathrm{arcsin}\frac{1}{2}-\mathrm{arcsin}\frac{1}{4}=\mathrm{arctan}\frac{1}{\sqrt{3}}-\mathrm{arctan}\frac{1}{\sqrt{15}}=\frac{\pi }{6}-\mathrm{arctan}\frac{1}{\sqrt{15}}$
which equals $\phantom{\rule{thickmathspace}{0ex}}\mathrm{arcsin}\frac{\sqrt{15}}{4}-\frac{\pi }{3}\phantom{\rule{thickmathspace}{0ex}}\dots \phantom{\rule{thickmathspace}{0ex}}$

Yahir Tucker

Expert

Notice that :
${\int }_{1/4}^{1/2}\frac{\mathrm{d}u}{\sqrt{1-{u}^{2}}}=-{\left[{\mathrm{cos}}^{-1}u\right]}_{1/4}^{1/2}={\mathrm{cos}}^{-1}\frac{1}{4}-{\mathrm{cos}}^{-1}\frac{1}{2}={\mathrm{cos}}^{-1}\frac{1}{4}-\frac{\pi }{3}$
But we know that :
${\mathrm{cos}}^{-1}x={\mathrm{sin}}^{-1}\sqrt{1-{x}^{2}}$
then :
${\int }_{1/4}^{1/2}\frac{\mathrm{d}u}{\sqrt{1-{u}^{2}}}={\mathrm{sin}}^{-1}\sqrt{1-\frac{1}{16}}-\frac{\pi }{3}={\mathrm{sin}}^{-1}\frac{\sqrt{15}}{4}-\frac{\pi }{3}$

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