`int(dx)/(3cos2x+5)=`

Options

A. `(1)/(2)tan^(-1)((tanx)/(2))+c`
B. `(1)/(2)tan^(-1)((tanx)/(4))+c`
C. `(1)/(4)tan^(-1)((tanx)/(2))+c`
D. `(1)/(4)tan^(-1)((tanx)/(4))+c`

Answer

To solve the integral \( I = \int \frac{dx}{3 \cos 2x + 5} \), we will follow these steps:

### Step 1: Use the identity for \( \cos 2x \)
We know that:
\[ \cos 2x = 2 \cos^2 x - 1 \]
Substituting this into the integral gives:
\[ I = \int \frac{dx}{3(2 \cos^2 x - 1) + 5} \]

### Step 2: Simplify the expression
Now, simplify the expression in the denominator:
\[ 3(2 \cos^2 x - 1) + 5 = 6 \cos^2 x - 3 + 5 = 6 \cos^2 x + 2 \]
Thus, we have:
\[ I = \int \frac{dx}{6 \cos^2 x + 2} \]

### Step 3: Factor out the common term
Factor out the 2 from the denominator:
\[ I = \int \frac{dx}{2(3 \cos^2 x + 1)} = \frac{1}{2} \int \frac{dx}{3 \cos^2 x + 1} \]

### Step 4: Multiply and divide by \( \sec^2 x \)
To facilitate integration, multiply and divide by \( \sec^2 x \):
\[ I = \frac{1}{2} \int \frac{\sec^2 x \, dx}{3 \cos^2 x \sec^2 x + \sec^2 x} = \frac{1}{2} \int \frac{\sec^2 x \, dx}{3 + \tan^2 x} \]

### Step 5: Use substitution
Let \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \). The integral becomes:
\[ I = \frac{1}{2} \int \frac{dt}{3 + t^2} \]

### Step 6: Recognize the integral form
The integral \( \int \frac{dt}{a^2 + t^2} \) has a standard result:
\[ \int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + C \]
In our case, \( a^2 = 3 \) so \( a = \sqrt{3} \):
\[ I = \frac{1}{2} \cdot \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{t}{\sqrt{3}} \right) + C \]

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

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

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