If `int ( dx )/(5 + 4 sin x) = A tan^(-1) ( B tan ((x)/(2)) + (4)/(3)) + C`, then :

Options

A. `A = (2)/(3) , B = (5)/(3)`
B. `A = (1)/(3), B = (5)/(3)`
C. `A = (2)/(3), B = (2)/(3)`
D. `A = (1)/(3), B = (2)/(3)`

Answer

To solve the integral \( \int \frac{dx}{5 + 4 \sin x} \) and express it in the form \( A \tan^{-1} \left( B \tan \left( \frac{x}{2} \right) + \frac{4}{3} \right) + C \), we will follow these steps:

### Step 1: Rewrite \( \sin x \) in terms of \( \tan \left( \frac{x}{2} \right) \)
Using the identity:
\[ \sin x = \frac{2 \tan \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \]
we can express \( 4 \sin x \) as:
\[ 4 \sin x = \frac{8 \tan \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)} \]

### Step 2: Substitute into the integral
Now, substituting this into the integral gives:
\[ \int \frac{dx}{5 + 4 \sin x} = \int \frac{dx}{5 + \frac{8 \tan \left( \frac{x}{2} \right)}{1 + \tan^2 \left( \frac{x}{2} \right)}} \]

### Step 3: Simplify the integral
Let \( t = \tan \left( \frac{x}{2} \right) \). Then, we have:
\[ dx = \frac{2}{1 + t^2} dt \]
Substituting \( dx \) into the integral:
\[ \int \frac{2 dt}{(5 + \frac{8t}{1+t^2})(1+t^2)} \]

### Step 4: Combine and simplify the denominator
The denominator becomes:
\[ 5(1 + t^2) + 8t = 5 + 5t^2 + 8t \]
Thus, we rewrite the integral as:
\[ \int \frac{2 dt}{5 + 5t^2 + 8t} \]

### Step 5: Factor the quadratic
The quadratic can be rewritten as:
\[ 5t^2 + 8t + 5 = 5 \left( t^2 + \frac{8}{5} t + 1 \right) \]
Completing the square:
\[ = 5 \left( \left( t + \frac{4}{5} \right)^2 + \frac{9}{25} \right) \]

### Step 6: Substitute back into the integral
Now, the integral becomes:
\[ \frac{2}{5} \int \frac{dt}{\left( t + \frac{4}{5} \right)^2 + \left( \frac{3}{5} \right)^2} \]

### Step 7: Use the arctangent formula
Using the formula:
\[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
we get:
\[ = \frac{2}{5} \cdot \frac{5}{3} \tan^{-1} \left( \frac{t + \frac{4}{5}}{\frac{3}{5}} \right) + C \]

### Step 8: Substitute back \( t = \tan \left( \frac{x}{2} \right) \)
Substituting \( t \) back:
\[ = \frac{2}{3} \tan^{-1} \left( \frac{5 \tan \left( \frac{x}{2} \right) + 4}{3} \right) + C \]

### Final Result
Thus, comparing with the form \( A \tan^{-1} \left( B \tan \left( \frac{x}{2} \right) + \frac{4}{3} \right) + C \), we find:
- \( A = \frac{2}{3} \)
- \( B = \frac{5}{3} \)