integration - Evaluating $int^{fracpi2}_{fracpi4}(2csc(x))^{17}dx$

I saw this question in JEE Advanced. But in that we had to simplify it to

$$\int^{\log(1+\sqrt2)}_{0}2(e^u+e^{-u})^{16}du$$
But I pose the following question as to evaluate this integral in closed form.
My Attempt:
$$\int^{\frac\pi2}_{\frac\pi4}(2\csc(x))^{17}dx$$
$$=\int^{\frac\pi2}_{\frac\pi4}\frac{(2.2i)^{17}}{(e^{ix}-e^{-ix})^{17}}dx$$
(Bit of a cheeky attempt! I know. :D)
$$4^{17}i\int^{\frac\pi2}_{\frac\pi4}\frac{e^{17ix}}{e^{2ix}-1}dx$$

If we put $u=e^{ix};du=i.e^{ix}dx$, then $$-4^{17}\int^{i}_{\frac{1+i}{\sqrt2}}\frac{u^{16}}{u^2-1}du$$
(Huh??!!)
$$4^{17}\int_i^{\frac{1+i}{\sqrt2}}\frac{u^{16}}{u^2-1}du$$
And $$\int\frac{u^{16}}{u^2-1}=\frac{u^{15}}{15}+\frac{u^{13}}{13}+\frac{u^{11}}{11}+\frac{u^{9}}{9}+\frac{u^{7}}{7}+\frac{u^{5}}{5}+\frac{u^{3}}{3}+u+\frac12\log(1-u)-\frac12\log(u+1)+C$$
But at this point I am stuck. Please can someone help?

Answer

By applying the Weierstrass substitution $x=2\arctan t$ we have:

$$\color{red}{I}=\int_{\pi/4}^{\pi/2}\left(\frac{2}{\sin x}\right)^{17}\,dx = \int_{\pi/4}^{\pi/2}\left(\frac{1}{\sin\frac{x}{2}\,\cos\frac{x}{2}}\right)^{17}=2\int_{\sqrt{2}-1}^{1}\left(\frac{1+t^2}{t}\right)^{17}\frac{dt}{1+t^2}$$
then by setting $t=e^{-u}$ and applying the binomial theorem it follows that:
$$\color{red}{I} = 2\int_{0}^{\log(1+\sqrt{2})}(e^u+e^{-u})^{16}\,du=\color{red}{\frac{16037316\,\sqrt{2}}{7}+25740\log\left(1+\sqrt{2}\right)}.$$