`int_(0)^(pi/4) (sec x)/(1+2 sin^(2)x)` is equal to

Options

A. `(1)/(3)[log(sqrt2+1)+(pi)/(2sqrt2)]`
B. `(1)/(3)[log(sqrt2+1)-(pi)/(2sqrt2)]`
C. `3[log(sqrt2+1)-(pi)/(2sqrt2)]`
D. `3[log(sqrt2+1)+(pi)/(2sqrt2)]`

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{4}} \frac{\sec x}{1 + 2 \sin^2 x} \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
We start by rewriting \(\sec x\) as \(\frac{1}{\cos x}\):
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos x (1 + 2 \sin^2 x)} \, dx \]

### Step 2: Multiply and divide by \(\cos x\)
Next, we multiply and divide the integrand by \(\cos x\):
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\cos x}{(1 + 2 \sin^2 x) \cos x} \, dx \]

### Step 3: Change of variable
Let \( t = \sin x \). Then, \( dt = \cos x \, dx \). When \( x = 0 \), \( t = 0 \) and when \( x = \frac{\pi}{4} \), \( t = \frac{1}{\sqrt{2}} \):
\[ I = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{(1 + 2t^2)(1 - t^2)} \, dt \]

### Step 4: Partial fraction decomposition
We need to decompose the integrand:
\[ \frac{1}{(1 - t^2)(1 + 2t^2)} = \frac{A}{1 - t^2} + \frac{B}{1 + 2t^2} \]
Multiplying through by the denominator:
\[ 1 = A(1 + 2t^2) + B(1 - t^2) \]
Expanding and rearranging gives:
\[ 1 = (A + B) + (2A - B)t^2 \]
By comparing coefficients, we get:
1. \( A + B = 1 \)
2. \( 2A - B = 0 \)

### Step 5: Solve for A and B
From \( 2A - B = 0 \), we have \( B = 2A \). Substituting into the first equation:
\[ A + 2A = 1 \implies 3A = 1 \implies A = \frac{1}{3} \]
Then, substituting back to find \( B \):
\[ B = 2A = \frac{2}{3} \]

### Step 6: Rewrite the integral
Now substituting \( A \) and \( B \) back into the integral:
\[ I = \int_{0}^{\frac{1}{\sqrt{2}}} \left( \frac{1/3}{1 - t^2} + \frac{2/3}{1 + 2t^2} \right) dt \]
This can be split into two separate integrals:
\[ I = \frac{1}{3} \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{1 - t^2} \, dt + \frac{2}{3} \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{1 + 2t^2} \, dt \]

### Step 7: Evaluate the integrals
1. The first integral:
\[ \int \frac{1}{1 - t^2} \, dt = \frac{1}{2} \log \left| \frac{1 + t}{1 - t} \right| + C \]
Evaluating from \( 0 \) to \( \frac{1}{\sqrt{2}} \):
\[ \frac{1}{2} \log \left( \frac{1 + \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} \right) \]

2. The second integral:
\[ \int \frac{1}{1 + 2t^2} \, dt = \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2} t) + C \]
Evaluating from \( 0 \) to \( \frac{1}{\sqrt{2}} \):
\[ \frac{1}{\sqrt{2}} \tan^{-1}(1) = \frac{\pi}{4\sqrt{2}} \]

### Step 8: Combine results
Combining both results:
\[ I = \frac{1}{3} \cdot \frac{1}{2} \log \left( \frac{1 + \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} \right) + \frac{2}{3} \cdot \frac{\pi}{4\sqrt{2}} \]

### Final Answer
After simplification, we find:
\[ I = \frac{1}{3} \cdot \frac{1}{2} \log \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right) + \frac{\pi}{6\sqrt{2}} \]