Let `I=int(dx)/(1+3sin^(2)x)=(1)/(2)tan^(-1)(2f(x))+C` (where, C is the constant of integration). If `f((pi)/(4))=1`, then the fundamental period of `y=f(x)` is

Options

A. `(pi)/(4)`
B. `pi`
C. `2pi`
D. `(pi)/(6)`

Answer

To solve the problem, we need to find the fundamental period of the function \( y = f(x) \) given the integral equation and the condition \( f\left(\frac{\pi}{4}\right) = 1 \).

### Step 1: Solve the integral

We start with the integral:

\[ I = \int \frac{dx}{1 + 3\sin^2 x} \]

To solve this integral, we can use the substitution \( \sin^2 x = 1 - \cos^2 x \) or directly apply the formula for integrals involving trigonometric functions. However, a more straightforward approach is to use the identity for \( \sin^2 x \):

\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]

This gives us:

\[ 1 + 3\sin^2 x = 1 + \frac{3}{2} - \frac{3}{2}\cos(2x) = \frac{5}{2} - \frac{3}{2}\cos(2x) \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{dx}{\frac{5}{2} - \frac{3}{2}\cos(2x)} = \frac{2}{5} \int \frac{dx}{1 - \frac{3}{5}\cos(2x)} \]

### Step 2: Use the integral formula

The integral \( \int \frac{dx}{a - b\cos(kx)} \) has a known solution:

\[ \int \frac{dx}{a - b\cos(kx)} = \frac{1}{\sqrt{a^2 - b^2}} \tan^{-1}\left(\frac{\sqrt{a^2 - b^2}}{a - b\cos(kx)}\right) + C \]

In our case, \( a = \frac{5}{2} \) and \( b = \frac{3}{2} \), and \( k = 2 \). So we compute:

\[ \sqrt{a^2 - b^2} = \sqrt{\left(\frac{5}{2}\right)^2 - \left(\frac{3}{2}\right)^2} = \sqrt{\frac{25}{4} - \frac{9}{4}} = \sqrt{\frac{16}{4}} = 2 \]

Thus, we have:

\[ I = \frac{2}{5} \cdot \frac{1}{2} \tan^{-1}\left(\frac{2}{\frac{5}{2} - \frac{3}{2}\cos(2x)}\right) + C = \frac{1}{5} \tan^{-1}\left(\frac{2}{\frac{5}{2} - \frac{3}{2}\cos(2x)}\right) + C \]

### Step 3: Relate to the given equation

We know from the problem statement that:

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

By comparing both expressions for \( I \), we can equate:

\[ \frac{1}{5} \tan^{-1}\left(\frac{2}{\frac{5}{2} - \frac{3}{2}\cos(2x)}\right) = \frac{1}{2} \tan^{-1}(2f(x)) + C \]

### Step 4: Find \( f(x) \)

From the equation above, we can solve for \( f(x) \). By rearranging, we find:

\[ 2f(x) = \frac{2}{\frac{5}{2} - \frac{3}{2}\cos(2x)} \]

Thus,

\[ f(x) = \frac{1}{\frac{5}{2} - \frac{3}{2}\cos(2x)} = \frac{2}{5 - 3\cos(2x)} \]

### Step 5: Determine the fundamental period

The function \( f(x) = \frac{2}{5 - 3\cos(2x)} \) is periodic. The cosine function \( \cos(2x) \) has a period of \( \frac{2\pi}{2} = \pi \). Therefore, the fundamental period of \( f(x) \) is:

\[ \text{Fundamental period of } f(x) = \pi \]

### Conclusion

The fundamental period of \( y = f(x) \) is:

\[ \boxed{\pi} \]