Q 57. - Calculus

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.

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

Step-by-Step Solution

Answer

The value of integral is $2$

Step 1: Given Information

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

Step 2: 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 $\cos x=t$, $\Rightarrow -{\csc }^{2}xdx=dt$

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

and

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

Hence,

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

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

$=2$