Integral Transformation of Higher Power Cosecant Function

To evaluate the definite integral $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 \csc x)^{17} \, dx$

we can use a substitution involving the standard trigonometric identity $\csc^2 x - \cot^2 x = 1$.

Let: $e^u = \csc x + \cot x$

Since $(\csc x + \cot x)(\csc x - \cot x) = \csc^2 x - \cot^2 x = 1$, we have: $e^{-u} = \frac{1}{\csc x + \cot x} = \csc x - \cot x$

Adding these two expressions gives: $e^u + e^{-u} = 2 \csc x$

Now, differentiating $u$ with respect to $x$: $u = \ln(\csc x + \cot x)$ $\frac{du}{dx} = \frac{-\csc x \cot x - \csc^2 x}{\csc x + \cot x} = \frac{-\csc x(\cot x + \csc x)}{\csc x + \cot x} = -\csc x$

Thus, the differential $dx$ can be written in terms of $u$ as: $dx = -\frac{du}{\csc x} = -\frac{2}{2\csc x} \, du = -\frac{2}{e^u + e^{-u}} \, du$

Next, we determine the new limits of integration:

• When $x = \frac{\pi}{4}$: $e^u = \csc\left(\frac{\pi}{4}\right) + \cot\left(\frac{\pi}{4}\right) = \sqrt{2} + 1 \implies u = \log(1 + \sqrt{2})$
• When $x = \frac{\pi}{2}$: $e^u = \csc\left(\frac{\pi}{2}\right) + \cot\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \implies u = \log(1) = 0$

Substituting $2\csc x$, $dx$, and the limits into the original integral: $I = \int_{\log(1+\sqrt{2})}^{0} (e^u + e^{-u})^{17} \left( -\frac{2}{e^u + e^{-u}} \right) du$

Using the negative sign to invert the limits of integration: $I = \int_{0}^{\log(1+\sqrt{2})} 2(e^u + e^{-u})^{16} \, du$

Therefore, the correct option is (A).