← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Riemann integral · Real Analysis · asked 10× in 13 yrs · Read the full method →

Question

Prove the inequality $\dfrac{\pi^2}{9}<\displaystyle\int_{\pi/6}^{\pi/2}\frac{x}{\sin x}\,dx<\dfrac{2\pi^2}{9}$.

Technique

Bracket the integrand by constants using strict monotonicity of $\sin x/x$ on $(0,\pi/2]$; integrate over the interval of length $\pi/3$.

Solution

The strategy is to bracket the integrand $\dfrac{x}{\sin x}$ between two constants on $[\pi/6,\pi/2]$ and integrate. The interval has length

Step 1 — Bracket the integrand using monotonicity of $\dfrac{\sin x}{x}$

Let $h(x)=\dfrac{\sin x}{x}$. On $(0,\pi/2]$,

because the numerator $\psi(x)=x\cos x-\sin x$ has $\psi(0)=0$ and $\psi'(x)=-x\sin x<0$, so $\psi(x)<0$ for $x\in(0,\pi/2]$. Hence $h$ is strictly decreasing, and on $[\pi/6,\pi/2]$,

Taking reciprocals (all quantities positive) reverses the inequalities:

with equality only at the endpoints, so the inequalities are strict on the open interval.

Step 2 — Lower bound

From the left half of (B), $\dfrac{x}{\sin x}>\dfrac{\pi}{3}$ on $(\pi/6,\pi/2)$. Integrating over the interval of length $\pi/3$:

Step 3 — Upper bound

From the right half of (B), $\dfrac{x}{\sin x}<\dfrac{\pi}{2}$ on $(\pi/6,\pi/2)$. Integrating:

Finally $\dfrac{\pi^2}{6}=\dfrac{3\pi^2}{18}<\dfrac{4\pi^2}{18}=\dfrac{2\pi^2}{9}$, so a fortiori

Conclusion

Answer