If `f(x)=int(x^(2)+sin^(2)x)/(1+x^(2))sec^(2)xdx and f(0)=0,` then `f(1)=`

Options

A. `1-(pi)/(4)`
B. `(pi)/(4)-1`
C. `(tan1)-(pi)/(4)`
D. `(pi)/(4)+tan1`

Answer

To solve the problem, we need to evaluate the integral given by:

\[ f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \sec^2 x \, dx \]

with the condition that \( f(0) = 0 \) and we need to find \( f(1) \).

### Step 1: Rewrite the integral

We start by rewriting the integrand:

\[ f(x) = \int \left( \frac{x^2}{1 + x^2} + \frac{\sin^2 x}{1 + x^2} \right) \sec^2 x \, dx \]

### Step 2: Split the integral

We can split the integral into two parts:

\[ f(x) = \int \frac{x^2}{1 + x^2} \sec^2 x \, dx + \int \frac{\sin^2 x}{1 + x^2} \sec^2 x \, dx \]

### Step 3: Simplify the first integral

The first integral can be simplified as follows:

\[ \int \frac{x^2}{1 + x^2} \sec^2 x \, dx = \int \left( 1 - \frac{1}{1 + x^2} \right) \sec^2 x \, dx \]

This gives us:

\[ \int \sec^2 x \, dx - \int \frac{\sec^2 x}{1 + x^2} \, dx \]

### Step 4: Evaluate the first integral

The integral of \( \sec^2 x \) is:

\[ \int \sec^2 x \, dx = \tan x \]

Thus, we have:

\[ \int \frac{x^2}{1 + x^2} \sec^2 x \, dx = \tan x - \int \frac{\sec^2 x}{1 + x^2} \, dx \]

### Step 5: Simplify the second integral

Now, we need to evaluate the second integral:

\[ \int \frac{\sin^2 x}{1 + x^2} \sec^2 x \, dx = \int \frac{\sin^2 x}{1 + x^2} \cdot \frac{1}{\cos^2 x} \, dx = \int \frac{\sin^2 x}{\cos^2 x (1 + x^2)} \, dx \]

### Step 6: Combine the results

Combining the results, we have:

\[ f(x) = \tan x - \int \frac{\sec^2 x}{1 + x^2} \, dx + \int \frac{\sin^2 x}{\cos^2 x (1 + x^2)} \, dx \]

### Step 7: Evaluate \( f(0) \)

We know that \( f(0) = 0 \). Substituting \( x = 0 \):

\[ f(0) = \tan(0) - \int \frac{\sec^2(0)}{1 + 0^2} \, dx + \int \frac{\sin^2(0)}{\cos^2(0)(1 + 0^2)} \, dx = 0 \]

This implies that the constant of integration \( C = 0 \).

### Step 8: Evaluate \( f(1) \)

Now, we need to find \( f(1) \):

\[ f(1) = \tan(1) - \int \frac{\sec^2(1)}{1 + 1^2} \, dx + \int \frac{\sin^2(1)}{\cos^2(1)(1 + 1^2)} \, dx \]

Calculating this gives us:

\[ f(1) = \tan(1) - \frac{1}{2} \sec^2(1) + \frac{\sin^2(1)}{2 \cos^2(1)} \]

### Final Result

Thus, the final value of \( f(1) \) is:

\[ f(1) = \tan(1) - \frac{1}{2} + \frac{\sin^2(1)}{2} \]