Find the area between the curves y = 1/(sin x) = csc x

Find the area between the curves $y = \frac{1}{\sin x} = \csc x$ and $y = \cot x$ from $x = 0$ to $x = \frac{\pi}{2}$.

We will show the difference of the two singular functions is itself perfectly well-behaved, then integrate it in closed form.

Step 1: Understand the curves

$y=\csc x=\frac{1}{\sin x}$ and $y=\cot x=\frac{\cos x}{\sin x}$ both individually blow up as $x\to0^+$ (each $\to+\infty$). As $x\to\frac\pi2^-$: $\csc x\to1$ and $\cot x\to0$.

Step 2: Determine which curve is above

\[\csc x - \cot x = \frac{1-\cos x}{\sin x}\]

For $x\in\left(0,\frac\pi2\right)$: $1-\cos x>0$ and $\sin x>0$, so $\csc x>\cot x$ throughout the interval.

Step 3: Show the difference is well-behaved at $x=0$

Using the half-angle identities $1-\cos x=2\sin^2\frac x2$ and $\sin x=2\sin\frac x2\cos\frac x2$:

\[\frac{1-\cos x}{\sin x} = \frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)} = \tan\frac x2\]

Since $\tan\frac{x}{2}\to0$ as $x\to0^+$, the integrand $\csc x-\cot x$ is actually finite (equal to $0$) at $x=0$ — even though each function individually diverges there, their difference does not. So the integral below is proper, not improper.

Step 4: Set up the integral

\[A = \int_0^{\pi/2} (\csc x - \cot x)\,dx = \int_0^{\pi/2} \tan\frac x2\,dx\]

Step 5: Find a closed form and integrate

Continuing from $\csc x-\cot x=\frac{1-\cos x}{\sin x}$, multiply and divide by $1+\cos x$:

\[\frac{1-\cos x}{\sin x} = \frac{(1-\cos x)(1+\cos x)}{\sin x(1+\cos x)} = \frac{\sin^2x}{\sin x(1+\cos x)} = \frac{\sin x}{1+\cos x}\]

and $\int \frac{\sin x}{1+\cos x}\,dx = -\ln(1+\cos x)+C$ (by the substitution $w=1+\cos x$). Therefore:

\[A = \left[-\ln(1+\cos x)\right]_0^{\pi/2}\]

At $x=\frac\pi2$: $-\ln(1+0)=0$. At $x=0$: $-\ln(1+1)=-\ln2$.

\[A = 0-(-\ln2) = \ln 2\]

\[\boxed{A = \ln 2 \text{ square units}}\]

Step 6: Numerical value

\[A = \ln 2 \approx 0.693 \text{ square units}\]

Summary

\[\boxed{A = \ln 2 \approx 0.693 \text{ square units}}\]

Key insight: Although $\csc x$ and $\cot x$ each blow up at $x=0$, their difference $\tan\frac{x}{2}$ stays finite there — a reminder that singularities of individual terms can cancel in a difference, turning an apparently improper integral into a perfectly ordinary one.