The following integral $\int_{\pi/4}^{\pi/2} (2\cosec x)^{17}\,dx$ is equal to

The following integral ∫_(π/4)^(π/2) (2cosec x)¹⁷ dx is equal to $I = \displaystyle\int_{\pi/4}^{\pi/2} (2\operatorname{cosec} x)^{17}\,dx$ $\text{To find: which option equals } I$ $\text{options use } u,\ \text{with } u:\ 0 \to \log(1+\sqrt{2}),\ \ (e^{u}\pm e^{-u})$ $\operatorname{cosec}^{2}x - \cot^{2}x = 1$ $u = \log(\operatorname{cosec} x + \cot x):\ e^{u} = \operatorname{cosec} x + \cot x,\ \ e^{-u} = \operatorname{cosec} x - \cot x$ $\Rightarrow\ e^{u} + e^{-u} = 2\operatorname{cosec} x$ $\dfrac{du}{dx} = \dfrac{-\operatorname{cosec} x\cot x - \operatorname{cosec}^{2}x}{\operatorname{cosec} x + \cot x} = -\operatorname{cosec} x$ $\Rightarrow\ dx = -\dfrac{du}{\operatorname{cosec} x}$ $(2\operatorname{cosec} x)^{17}dx = (e^{u}+e^{-u})^{17}\cdot\left(-\dfrac{2\,du}{e^{u}+e^{-u}}\right)$ $= -2\,(e^{u}+e^{-u})^{16}\,du$ $x=\dfrac{\pi}{4} \Rightarrow u = \log(\sqrt{2}+1); \qquad x=\dfrac{\pi}{2} \Rightarrow u = \log 1 = 0$ $I = \displaystyle\int_{\log(\sqrt{2}+1)}^{0} -2\,(e^{u}+e^{-u})^{16}\,du = \displaystyle\int_{0}^{\log(1+\sqrt{2})} 2\,(e^{u}+e^{-u})^{16}\,du$ $\boxed{\text{Answer: (1)}}$