Let `f(x)=1/(4-3cos^2x+5sin^2x)` and if its antiderivative `F(x)=(1/3) tan^-1(g(x))+C` then `g(x)` is equal to

Options

A. `3 tan x`
B. `sqrt2 tan x`
C. `2 tan x`
D. None of these

Answer

To solve the problem, we need to find the function \( g(x) \) such that the antiderivative \( F(x) \) of the function \( f(x) = \frac{1}{4 - 3\cos^2 x + 5\sin^2 x} \) can be expressed as \( F(x) = \frac{1}{3} \tan^{-1}(g(x)) + C \).

### Step-by-Step Solution:

1. **Rewrite the Function \( f(x) \)**:
\[ f(x) = \frac{1}{4 - 3\cos^2 x + 5\sin^2 x} \]
We can use the identity \( \sin^2 x + \cos^2 x = 1 \) to rewrite \( f(x) \):
\[ f(x) = \frac{1}{4 - 3\cos^2 x + 5(1 - \cos^2 x)} = \frac{1}{4 - 3\cos^2 x + 5 - 5\cos^2 x} = \frac{1}{9 - 8\cos^2 x} \]

2. **Express in Terms of \( \tan x \)**:
We can express \( \cos^2 x \) in terms of \( \tan x \):
\[ \cos^2 x = \frac{1}{1 + \tan^2 x} \]
Thus,
\[ f(x) = \frac{1}{9 - 8\left(\frac{1}{1 + \tan^2 x}\right)} = \frac{1 + \tan^2 x}{9(1 + \tan^2 x) - 8} \]
Simplifying gives:
\[ f(x) = \frac{1 + \tan^2 x}{9 + 9\tan^2 x - 8} = \frac{1 + \tan^2 x}{1 + 9\tan^2 x} \]

3. **Integrate \( f(x) \)**:
Now, we need to find the integral:
\[ F(x) = \int f(x) \, dx = \int \frac{1 + \tan^2 x}{1 + 9\tan^2 x} \, dx \]
We can split the integral:
\[ F(x) = \int \frac{1}{1 + 9\tan^2 x} \, dx + \int \frac{\tan^2 x}{1 + 9\tan^2 x} \, dx \]

4. **Use Substitution for the First Integral**:
For the first integral, we can use the formula:
\[ \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \]
Here, \( a^2 = \frac{1}{9} \) so \( a = \frac{1}{3} \):
\[ \int \frac{dx}{1 + 9\tan^2 x} = \frac{1}{3} \tan^{-1}(3\tan x) + C \]

5. **Combine Results**:
Thus, we have:
\[ F(x) = \frac{1}{3} \tan^{-1}(3\tan x) + C \]

6. **Identify \( g(x) \)**:
From the expression \( F(x) = \frac{1}{3} \tan^{-1}(g(x)) + C \), we can see that:
\[ g(x) = 3\tan x \]

### Final Answer:
\[ g(x) = 3\tan x \]