The user wants me to solve the definite integral: $I = \int\limits_{{\pi / 4}}^{{\pi / 2}} {{{\left( {2\csc\,x} \right)}^{17}}dx}$ and match it with one of the given options. The options are integrals in terms of a variable u with exponential functions, which suggests that a substitution is required to transform the trigonometric integral into one with exponential functions.

Step-by-step solution:

1. Choosing the substitution: The options have limits of integration from $0$ to $\log(1+\sqrt{2})$ and involve terms like $e^u + e^{-u}$. This hints at a standard substitution used for integrals involving $\csc x$. Let's try the substitution: $u = \log(\csc x + \cot x)$

2. Changing the limits of integration: We need to find the new limits for the variable $u$ corresponding to the original limits for $x$.

   • Lower limit: When $x = \pi/4$, $u = \log(\csc(\pi/4) + \cot(\pi/4)) = \log(\sqrt{2} + 1)$
   • Upper limit: When $x = \pi/2$, $u = \log(\csc(\pi/2) + \cot(\pi/2)) = \log(1 + 0) = \log(1) = 0$ So, the integration will be from $\log(1+\sqrt{2})$ to $0$.

3. Expressing the integrand in terms of u: From our substitution, we have $e^u = \csc x + \cot x$. Then, $e^{-u} = \frac{1}{\csc x + \cot x}$. Rationalizing the denominator gives: $e^{-u} = \frac{1}{\csc x + \cot x} \times \frac{\csc x - \cot x}{\csc x - \cot x} = \frac{\csc x - \cot x}{\csc^2 x - \cot^2 x} = \csc x - \cot x$ Now, we can find an expression for $\csc x$ by adding $e^u$ and $e^{-u}$: $e^u + e^{-u} = (\csc x + \cot x) + (\csc x - \cot x) = 2\csc x$ The term in the integrand is $(2\csc x)^{17}$, which becomes: $(2\csc x)^{17} = (e^u + e^{-u})^{17}$

4. Finding dx in terms of du: We differentiate the substitution $u = \log(\csc x + \cot x)$ with respect to $x$: $\frac{du}{dx} = \frac{1}{\csc x + \cot x} \cdot \frac{d}{dx}(\csc x + \cot x)$ $\frac{du}{dx} = \frac{1}{\csc x + \cot x} \cdot (-\csc x \cot x - \csc^2 x)$ $\frac{du}{dx} = \frac{-\csc x (\cot x + \csc x)}{\csc x + \cot x} = -\csc x$ So, $du = -\csc x \, dx$. This gives $dx = -\frac{du}{\csc x}$. From step 3, we know $\csc x = \frac{e^u + e^{-u}}{2}$. Substituting this into the expression for $dx$: $dx = -\frac{du}{(\frac{e^u + e^{-u}}{2})} = -\frac{2 \, du}{e^u + e^{-u}}$

5. Substituting into the integral and simplifying: Now we substitute the new limits, the new integrand, and the new differential into the original integral: $I = \int\limits_{\log(1+\sqrt{2})}^{0} (e^u + e^{-u})^{17} \left( -\frac{2 \, du}{e^u + e^{-u}} \right)$ $I = -2 \int\limits_{\log(1+\sqrt{2})}^{0} \frac{(e^u + e^{-u})^{17}}{e^u + e^{-u}} \, du$ $I = -2 \int\limits_{\log(1+\sqrt{2})}^{0} (e^u + e^{-u})^{16} \, du$ Using the property of definite integrals $\int_a^b f(x)dx = -\int_b^a f(x)dx$, we can flip the limits of integration and remove the negative sign: $I = 2 \int\limits_{0}^{\log(1+\sqrt{2})} (e^u + e^{-u})^{16} \, du$

6. Comparing with the options: The calculated integral is: $I = \int\limits_0^{\log \left( {1 + \sqrt 2 } \right)} {2{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}\,du}$ This matches option A exactly.

Conclusion

The correct option is A.