Question

If \[ \int_0^\pi \frac{dx}{1+2\sin^2x}=k, \] then greatest integer less than or equal to \(k\) is

• A. \(2\)
• B. \(0\)
• C. \(1\)
• D. \(-1\)

Hint:

For integrals of the form \(\int_0^{\pi/2}\frac{dx}{a+b\sin^2x}\), directly use \(\frac{\pi}{2\sqrt{a(a+b)}}\).

Answer 1

The Correct Option is C

Step 1: Use symmetry of the integral.
We have \[ k=\int_0^\pi \frac{dx}{1+2\sin^2x} \] Since \[ \sin(\pi-x)=\sin x, \] we can write \[ k=2\int_0^{\pi/2}\frac{dx}{1+2\sin^2x} \]

Step 2: Use the standard formula.
The standard result is \[ \int_0^{\pi/2}\frac{dx}{a+b\sin^2x} = \frac{\pi}{2\sqrt{a(a+b)}} \] Here, \[ a=1,\quad b=2 \] So, \[ \int_0^{\pi/2}\frac{dx}{1+2\sin^2x} = \frac{\pi}{2\sqrt{1(1+2)}} \] \[ = \frac{\pi}{2\sqrt{3}} \] Therefore, \[ k=2\cdot \frac{\pi}{2\sqrt{3}} \] \[ k=\frac{\pi}{\sqrt{3}} \]

Step 3: Find the greatest integer less than or equal to \(k\).
Since \[ \pi\approx 3.14 \] and \[ \sqrt{3}\approx 1.732, \] we get \[ k=\frac{\pi}{\sqrt{3}}\approx 1.81 \] Therefore, \[ \lfloor k\rfloor=1 \]

Step 4: Final conclusion.
Hence, \[ \boxed{1} \]