Question

Sketch the region determined by the iterated integral and then evaluate the integral. For some of these integrals, it may be helpful to reverse the order of integration.

${\int }_{\pi /4}^{3\pi /4}{\int }_0^{\csc y}\csc ydxdy$

Short answer

The value of integral is$2$

Step-by-step solution

Given Information

It is given that ${\int }_{\pi /4}^{3\pi /4}{\int }_0^{\csc y}\csc ydxdy$

Simplification

The region of integration bounded by curve $y={\csc }^{-1}x$, line $x=0$, $y=\frac{\pi }{4}$and $y=\frac{3\pi }{4}$

Integral is simplified as

${\int }_{\pi /4}^{3\pi /4}{\int }_0^{\csc y}\csc ydxdy={\int }_{\pi /4}^{3\pi /4}[x{]}_0^{sscy}\csc ydy$

$={\int }_{\pi /4}^{3\pi /4}{\csc }^{2}ydy$

Using $\cos x=t$, role="math" localid="1653934317274" $\Rightarrow -{\csc }^{2}xdx=dt$

Limits are $t=\cot \frac{\pi }{4}=1$

and

$t=\cot \frac{3\pi }{4}=-1$

Hence,

${\int }_{\pi /4}^{3\pi /4}{\int }_0^{escy}\csc ydxdy=-{\int }_1^{-1}dt$

$=-[t{]}_1^{-1}$

$=2$