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

\[ I = \int \frac{dx}{5 + 4 \sin x} \]

we will follow these steps:

### Step 1: Rewrite the Integral
We can use the identity for sine in terms of tangent:

\[ \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]

Substituting this into the integral gives:

\[ I = \int \frac{dx}{5 + 4 \left(\frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\right)} \]

### Step 2: Simplify the Expression
This can be rewritten as:

\[ I = \int \frac{dx}{5 + \frac{8 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} \]

To combine the terms, we multiply the numerator and denominator by \(1 + \tan^2 \frac{x}{2}\):

\[ I = \int \frac{(1 + \tan^2 \frac{x}{2}) \, dx}{5(1 + \tan^2 \frac{x}{2}) + 8 \tan \frac{x}{2}} \]

### Step 3: Change of Variables
Let \(t = \tan \frac{x}{2}\). Then, we have:

\[ dx = \frac{2}{1 + t^2} dt \]

Substituting this in gives:

\[ I = \int \frac{(1 + t^2) \cdot \frac{2}{1 + t^2} dt}{5(1 + t^2) + 8t} = \int \frac{2 \, dt}{5(1 + t^2) + 8t} \]

### Step 4: Factor the Denominator
The denominator can be rearranged:

\[ 5(1 + t^2) + 8t = 5t^2 + 8t + 5 \]

### Step 5: Complete the Square
To integrate, we complete the square:

\[ 5t^2 + 8t + 5 = 5\left(t^2 + \frac{8}{5}t + 1\right) = 5\left(\left(t + \frac{4}{5}\right)^2 + \frac{9}{25}\right) \]

### Step 6: Substitute Back into the Integral
Now we have:

\[ I = \frac{2}{5} \int \frac{dt}{\left(t + \frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2} \]

This integral is in the standard form:

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

### Step 7: Solve the Integral
Thus, we get:

\[ I = \frac{2}{5} \cdot \frac{1}{\frac{3}{5}} \tan^{-1} \left(\frac{t + \frac{4}{5}}{\frac{3}{5}}\right) + C = \frac{2}{3} \tan^{-1} \left(\frac{5t + 4}{3}\right) + C \]

### Step 8: Substitute Back for \(t\)
Substituting back \(t = \tan \frac{x}{2}\):

\[ I = \frac{2}{3} \tan^{-1} \left(5 \tan \frac{x}{2} + \frac{4}{3}\right) + C \]

### Step 9: Compare with Given Form
We compare this with the given form:

\[ A \tan^{-1} \left(B \tan \frac{x}{2} + \frac{4}{3}\right) + C \]

From this comparison, we find:

\[ A = \frac{2}{3}, \quad B = 5 \]

### Final Result
Thus, the values of \(A\) and \(B\) are:

\[ A = \frac{2}{3}, \quad B = 5 \]

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