Question

Evaluate the iterated integrals in Exercises 45–48 by reversing the order of integration. Explain why it is easier to reverse the order of integration than evaluate the given iterated integral.

${\int }_0^1{\int }_{\frac{\mathrm{\pi }}{4}}^{co{t}^{-1}y}cscx\mathrm{dx}\mathrm{dy}$

Short answer

The value of given integral is:$\sqrt{2}-1$

The function$cscx$has a simpler antiderivative when integrated with respect to$y$than it does when integrated with respect to$x$.

Step-by-step solution

Step 1. Given Information

An integral,${\int }_0^1{\int }_{\frac{\mathrm{\pi }}{4}}^{co{t}^{-1}y}cscx\mathrm{dx}\mathrm{dy}$

Step 2. Evaluating the given iterated integrals by reversing the order of integration.

Given integral,

${\int }_0^1{\int }_{\frac{\mathrm{\pi }}{4}}^{co{t}^{-1}y}cscx\mathrm{dx}\mathrm{dy}$

Reversing the order of integration,

$$\begin{gathered} {\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}{\int }_0^{cotx}cscxdydx \\ ={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\left(cscx\right){\overline{)y}}_0^{cotx}dx \\ ={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\left(cscx\right)(cotx)dx \\ ={\overline{)-cscx}}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}} \\ =-\left(csc\frac{\mathrm{\pi }}{2}-csc\frac{\mathrm{\pi }}{4}\right) \\ =\sqrt{2}-1 \end{gathered}$$

The function$cscx$has a simpler antiderivative when integrated with respect to$y$than it does when integrated with respect to$x$.