The integral ∫limits_π / 4^π / 2(2 cosec x)^17 d x is equal to

Let $ I=\int\limits_{\pi / 4}^{\pi / 2}(2 cosec~ x)^{17} d x$ and let $e^u+e^{-u}=2 cosec~ x$ Then,

$x=\frac{\pi}{4} \Rightarrow e^u+e^{-u}=2 \sqrt{2}$

$\Rightarrow \left(e^u\right)^2-2 \sqrt{2} e^u+1=0$

$\Rightarrow e^u=\sqrt{2}+1 \Rightarrow u=\log (1+\sqrt{2})$

$x=\frac{\pi}{2} \Rightarrow e^u+e^{-u}=2$

$\Rightarrow \left(e^u-1\right)^2=0 \Rightarrow e^u=1 \Rightarrow u=0$

Now, $e^u+e^{-u}=2 cosec~ x$

$\Rightarrow \left(e^u\right)^2-(2 cosec~ x) e^u+1=0$

$\Rightarrow e^u=cosec~ x+\cot x$

$\Rightarrow e^{-u}=\frac{1}{cosec~ x+\cot x}=cosec~ x-\cot x$

∴  $\cot x=\frac{e^u-e^{-u}}{2}$

Again, $e^u+e^{-u}=2 cosec~ x$

$\Rightarrow \left(e^u-e^{-u}\right) d u=-2 cosec~x \cot x d x$

$\Rightarrow d x=-\frac{\left(e^u-e^{-u}\right)}{2 cosec~x \cot x} d u$

$\Rightarrow d x=-\frac{\left(e^u-e^{-u}\right)}{\left(e^u+e^{-u}\right)\left(\frac{e^u-e^{-u}}{2}\right)} d u$

$\Rightarrow d x=\frac{-2}{e^u+e^{-u}} d u$

∴  $I=\int\limits_{\pi / 4}^{\pi / 2}(2 cosec~ x)^{17} d x$

$\Rightarrow I=\int\limits_{\log (1+\sqrt{2})}^0\left(e^u+e^{-u}\right)^{17} \times \frac{-2}{\left(e^u+e^{-u}\right)} d u$

$\Rightarrow I=\int\limits_0^{\log (1+\sqrt{2})} 2\left(e^u+e^{-u}\right)^{16} d u$