We are asked to evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \frac{1}{\sin^2 x + 3 \cos^2 x} dx$.

1. Problem Description

We are asked to evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \frac{1}{\sin^2 x + 3 \cos^2 x} dx$.

2. Solution Steps

First, we divide both the numerator and the denominator by $\cos^2 x$.

$$\int_{0}^{\frac{\pi}{4}} \frac{1}{\sin^2 x + 3 \cos^2 x} dx = \int_{0}^{\frac{\pi}{4}} \frac{\frac{1}{\cos^2 x}}{\frac{\sin^2 x}{\cos^2 x} + 3 \frac{\cos^2 x}{\cos^2 x}} dx$$

This simplifies to:

$$\int_{0}^{\frac{\pi}{4}} \frac{\sec^2 x}{\tan^2 x + 3} dx$$

Let $u = \tan x$. Then, $du = \sec^2 x dx$. The limits of integration will change: when $x = 0$, $u = \tan 0 = 0$; when $x = \frac{\pi}{4}$, $u = \tan \frac{\pi}{4} = 1$.

Thus, the integral becomes:

$$\int_{0}^{1} \frac{1}{u^2 + 3} du$$

We know that

$$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \frac{x}{a} + C$$

So, in our case, $a^2 = 3$, so $a = \sqrt{3}$. Therefore,

$$\int_{0}^{1} \frac{1}{u^2 + 3} du = \left[ \frac{1}{\sqrt{3}} \arctan \frac{u}{\sqrt{3}} \right]_{0}^{1} = \frac{1}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3}} \arctan 0$$

Since $\arctan 0 = 0$, we have

$$\frac{1}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}}$$

We know that $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$. Therefore, $\arctan \frac{1}{\sqrt{3}} = \frac{\pi}{6}$.

So, the integral evaluates to:

$$\frac{1}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{6\sqrt{3}}$$

Rationalizing the denominator gives us:

$$\frac{\pi \sqrt{3}}{6 \cdot 3} = \frac{\pi \sqrt{3}}{18}$$

3. Final Answer

$\frac{\pi \sqrt{3}}{18}$