To evaluate the integral
\[
I = \int \frac{\ln(1 + \sin^2 x)}{\cos^2 x} \, dx,
\]
we will use integration by parts. Let's denote:
- \( u = \ln(1 + \sin^2 x) \)
- \( dv = \sec^2 x \, dx \)
### Step 1: Differentiate \( u \) and integrate \( dv \)
First, we need to find \( du \) and \( v \):
1. Differentiate \( u \):
\[
du = \frac{d}{dx} \ln(1 + \sin^2 x) \, dx = \frac{2 \sin x \cos x}{1 + \sin^2 x} \, dx = \frac{\sin(2x)}{1 + \sin^2 x} \, dx.
\]
2. Integrate \( dv \):
\[
v = \int \sec^2 x \, dx = \tan x.
\]
### Step 2: Apply Integration by Parts Formula
Now, we apply the integration by parts formula:
\[
\int u \, dv = uv - \int v \, du.
\]
Substituting our values:
\[
I = \ln(1 + \sin^2 x) \tan x - \int \tan x \cdot \frac{\sin(2x)}{1 + \sin^2 x} \, dx.
\]
### Step 3: Simplify the Integral
The integral we need to evaluate now is:
\[
\int \tan x \cdot \frac{\sin(2x)}{1 + \sin^2 x} \, dx.
\]
Recall that \( \sin(2x) = 2 \sin x \cos x \), so we can rewrite the integral as:
\[
\int \tan x \cdot \frac{2 \sin x \cos x}{1 + \sin^2 x} \, dx = 2 \int \frac{\sin^2 x}{1 + \sin^2 x} \, dx.
\]
### Step 4: Split the Integral
Now, we can split the integral:
\[
\int \frac{\sin^2 x}{1 + \sin^2 x} \, dx = \int \left( 1 - \frac{1}{1 + \sin^2 x} \right) \, dx.
\]
This gives us:
\[
\int dx - \int \frac{1}{1 + \sin^2 x} \, dx.
\]
### Step 5: Evaluate Each Integral
1. The first integral is straightforward:
\[
\int dx = x.
\]
2. The second integral can be evaluated using the identity:
\[
\int \frac{1}{1 + \sin^2 x} \, dx = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{\sin x}{\sqrt{3}} \right) + C.
\]
### Step 6: Combine Everything
Now we can combine everything back into our expression for \( I \):
\[
I = \ln(1 + \sin^2 x) \tan x - 2 \left( x - \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{\sin x}{\sqrt{3}} \right) \right) + C.
\]
Thus, the final result is:
\[
I = \ln(1 + \sin^2 x) \tan x - 2x + \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{\sin x}{\sqrt{3}} \right) + C.
\]
