$$\text { Let } \mathrm{I}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2 \operatorname{cosec} x)^{17} \cdot d x \quad \text{... (i)}$$

Let, $$\operatorname{cosec} x+\cot x=e^u\quad \text{... (ii)}$$

We know that $$\quad \operatorname{cosec}^2 x-\cot ^2 x=1$$

$$\begin{aligned} \Rightarrow \quad& (\operatorname{cosec} x+\cot x)(\operatorname{cosec} x-\cot x) =1 \\ \Rightarrow \quad& e^u(\operatorname{cosec} x-\cot x) =1 \\ \Rightarrow \quad& \operatorname{cosec} x-\cot x =e^{-u} \quad \text{... (iii)} \end{aligned}$$

From equation (ii) and (iii)

$$\begin{array}{rlrl} \Rightarrow & 2 \cot x =e^{+u}-e^{-u} \quad \text{... (iv)}\\ \text { And } & 2 \operatorname{cosec} x =e^u+e^{-u} \quad \text{... (v)} \end{array}$$

Differentiate the equation $$(v)$$ w.r.t. $$x$$

$$\begin{array}{llrl} \Rightarrow & -2 \operatorname{cosec} x \cdot \cot x =\left(e^u-e^{-u}\right) \frac{d u}{d x} \\ \Rightarrow & -\left(e^u+e^{-u}\right) \cdot\left(\frac{e^u-e^{-u}}{2}\right) =\left(e^u-e^{-u}\right) \frac{d u}{d x} \\ \Rightarrow & d x =-\frac{2 d u}{e^u+e^{-u}} \end{array}$$

Now, the equation (i) can be written as

$$\begin{aligned} & \Rightarrow \mathrm{I}=\int_{\ln (\sqrt{2}+1)}^{\ln (1+0)}\left(e^u+e^{-u}\right)^{17}\left(-\frac{2 d u}{e^u+e^{-u}}\right) \\ & \Rightarrow \mathrm{I}=-\int_{\ln (\sqrt{2}+1)}^0 2\left(e^u+e^{-u}\right)^{16} d u \\ & \Rightarrow \mathrm{I}=\int_0^{\ln (\sqrt{2}+1)} 2\left(e^u+e^{-u}\right)^{16} d u \end{aligned}$$

Hint:

(i) Recall $$\operatorname{cosec}^2 x-\cot ^2 x=1$$

(ii) If $$\operatorname{cosec} x+\cot x=e^u$$, then $$\operatorname{cosec} x-\cot x=e^{-u}, 2 \operatorname{cosec} x=e^u+e^{-u}, 2 \cot x=e^u-e^{-u}$$.