What is `int_0^pi (dx)/(1+2sin^2x) ` equal to

Options

A. `pi`
B. `pi/3`
C. `pi/sqrt3`
D. `(2 pi)/sqrt3`

Answer

To solve the integral \( I = \int_0^\pi \frac{dx}{1 + 2 \sin^2 x} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_0^\pi \frac{dx}{1 + 2 \sin^2 x} \]

### Step 2: Use a Trigonometric Identity
We can use the identity \( \sin^2 x = 1 - \cos^2 x \) to rewrite the integral:
\[ I = \int_0^\pi \frac{dx}{1 + 2(1 - \cos^2 x)} = \int_0^\pi \frac{dx}{3 - 2\cos^2 x} \]

### Step 3: Multiply by \(\cos^2 x\)
To simplify the integral, we multiply the numerator and the denominator by \(\cos^2 x\):
\[ I = \int_0^\pi \frac{\cos^2 x \, dx}{\cos^2 x(3 - 2\cos^2 x)} = \int_0^\pi \frac{\cos^2 x \, dx}{3\cos^2 x - 2\cos^4 x} \]

### Step 4: Use Substitution
Let \( t = \cot x \), which implies \( dx = -\frac{dt}{1 + t^2} \). The limits change as follows: when \( x = 0 \), \( t \to \infty \) and when \( x = \pi \), \( t \to -\infty \). Thus, we have:
\[ I = \int_{\infty}^{-\infty} \frac{\frac{1}{t^2 + 1}}{3\frac{1}{t^2} - 2\frac{1}{t^4}} (-dt) \]

### Step 5: Simplify the Integral
This becomes:
\[ I = \int_{-\infty}^{\infty} \frac{dt}{3 + 2t^2} \]

### Step 6: Evaluate the Integral
The integral can be evaluated using the formula for the integral of the form \( \int \frac{dx}{a + bx^2} \):
\[ I = \frac{\pi}{\sqrt{3 \cdot 2}} = \frac{\pi}{\sqrt{6}} \]

### Step 7: Final Result
Thus, the final result is:
\[ I = \frac{\pi}{\sqrt{3}} \]