1 <= (1)/(sin x) <= 2″ for all “x in[(pi)/(6),(pi)/(2)]”. “1 \leq \frac{1}{\sin x} \leq 2 \text { for all } x \in\left[\frac{\pi}{6}, \frac{\pi}{2}\right] \text {. }
Multiply by xx
Therefore, it follows that:
x <= (x)/(sin x) <= 2x” for all “x in[(pi)/(6),(pi)/(2)]x \leq \frac{x}{\sin x} \leq 2x \text { for all } x \in\left[\frac{\pi}{6}, \frac{\pi}{2}\right]
Step 1: Defining f(x)f(x):
Let’s define a function f(x)=(x)/(sin x)f(x)=\frac{x}{\sin x}.
Step 2: Defining phi(x)\phi(x) and psi(x)\psi(x):
We define two more functions as follows:
{:[phi(x)=x],[psi(x)=2x”,”quad x in[pi//6″,”pi//2]]:}\begin{aligned}
& \phi(x)=x \\
& \psi(x)=2x, \quad x \in[\pi / 6, \pi / 2]
\end{aligned}
Step 3: Boundedness and Integrability:
Both ff and phi\phi are bounded and integrable on [pi//6,pi//2][\pi / 6, \pi / 2], and f(x) >= phi(x)f(x) \geq \phi(x) for all x in[pi//6,pi//2]x \in[\pi / 6, \pi / 2].
Step 4: Continuity at pi//3\pi / 3:
Furthermore, ff and phi\phi are both continuous at x=pi//3x = \pi / 3, and f(pi//3) > phi(pi//3)f(\pi / 3)>\phi(\pi / 3).
Step 5: Integral Comparison:
Hence, we can compare the integrals:
{:[int_(pi//6)^(pi//2)f(x)dx > int_(pi//6)^(pi//2)phi(x)dx],[=int_(pi//6)^(pi//2)xdx],[=(pi^(2))/(9)]:}\begin{aligned}
& \int_{\pi / 6}^{\pi / 2} f(x) d x>\int_{\pi / 6}^{\pi / 2} \phi(x) d x \\
& =\int_{\pi / 6}^{\pi / 2} x d x \\
& =\frac{\pi^2}{9}
\end{aligned}
Step 6: Comparing with psi\psi:
Similarly, we have:
{:[int_(pi//6)^(pi//2)f(x)dx < int_(pi//6)^(pi//2)psi(x)dx],[=2int_(pi//6)^(pi//2)xdx],[=(2pi^(2))/(9)]:}\begin{aligned}
\int_{\pi / 6}^{\pi / 2} f(x) d x & <\int_{\pi / 6}^{\pi / 2} \psi(x) d x \\
& =2 \int_{\pi / 6}^{\pi / 2} x d x \\
& =\frac{2 \pi^2}{9}
\end{aligned}