`int(dx)/(4+5cos^2x)=`

Options

A. `(1)/(6)tan^(-1)((2tanx)/(3))+c`
B. `(1)/(3)tan^(-1)((2tanx)/(3))+c`
C. `(2)/(3)tan^(-1)((2tanx)/(3))+c`
D. `(1)/(12)tan^(-1)((2tanx)/(3))+c`

Answer

To solve the integral \( I = \int \frac{dx}{4 + 5 \cos^2 x} \), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{dx}{4 + 5 \cos^2 x} \]

### Step 2: Use the identity for \(\cos^2 x\)
We can express \(\cos^2 x\) in terms of \(\tan^2 x\):
\[ \cos^2 x = \frac{1}{1 + \tan^2 x} \]
Thus, we can rewrite the integral:
\[ I = \int \frac{dx}{4 + 5 \cdot \frac{1}{1 + \tan^2 x}} = \int \frac{dx}{4 + \frac{5}{1 + \tan^2 x}} \]

### Step 3: Simplify the denominator
Multiplying the numerator and denominator by \(1 + \tan^2 x\):
\[ I = \int \frac{(1 + \tan^2 x) \, dx}{(4(1 + \tan^2 x) + 5)} \]
This simplifies to:
\[ I = \int \frac{(1 + \tan^2 x) \, dx}{4 + 5 + 4 \tan^2 x} = \int \frac{(1 + \tan^2 x) \, dx}{9 + 4 \tan^2 x} \]

### Step 4: Substitute \( \tan x = y \)
Let \( \tan x = y \), then \( dx = \frac{dy}{1 + y^2} \):
\[ I = \int \frac{(1 + y^2) \cdot \frac{dy}{1 + y^2}}{9 + 4y^2} = \int \frac{dy}{9 + 4y^2} \]

### Step 5: Factor out constants
We can factor out the constant in the denominator:
\[ I = \frac{1}{4} \int \frac{dy}{\frac{9}{4} + y^2} \]

### Step 6: Use the standard integral formula
The standard integral formula is:
\[ \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
In our case, \( a^2 = \frac{9}{4} \) so \( a = \frac{3}{2} \):
\[ I = \frac{1}{4} \cdot \frac{1}{\frac{3}{2}} \tan^{-1} \left( \frac{y}{\frac{3}{2}} \right) + C \]

### Step 7: Substitute back \( y = \tan x \)
Substituting back \( y = \tan x \):
\[ I = \frac{1}{6} \tan^{-1} \left( \frac{2}{3} \tan x \right) + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = \frac{1}{6} \tan^{-1} \left( \frac{2}{3} \tan x \right) + C \]