`int(1)/(5-4 sin x)dx`

Answer

To solve the integral \( I = \int \frac{1}{5 - 4 \sin x} \, dx \), we will use a substitution method along with some trigonometric identities.

### Step 1: Substitute for \(\sin x\)
We can use the identity for \(\sin x\) in terms of tangent:
\[ \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]
Let \( t = \tan \frac{x}{2} \). Then, we have:
\[ \sin x = \frac{2t}{1 + t^2} \]

### Step 2: Rewrite the integral
Substituting \(\sin x\) into the integral gives:
\[ I = \int \frac{1}{5 - 4 \left( \frac{2t}{1 + t^2} \right)} \, dx \]
This simplifies to:
\[ I = \int \frac{1}{5 - \frac{8t}{1 + t^2}} \, dx \]

### Step 3: Simplify the denominator
To simplify the denominator, multiply both the numerator and the denominator by \(1 + t^2\):
\[ I = \int \frac{1 + t^2}{5(1 + t^2) - 8t} \, dx \]

### Step 4: Change of variable for \(dx\)
Using the substitution \( t = \tan \frac{x}{2} \), we have:
\[ dx = \frac{2}{1 + t^2} \, dt \]
Substituting this into the integral gives:
\[ I = \int \frac{1 + t^2}{5(1 + t^2) - 8t} \cdot \frac{2}{1 + t^2} \, dt \]
This simplifies to:
\[ I = 2 \int \frac{1}{5(1 + t^2) - 8t} \, dt \]

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

### Step 6: Substitute back into the integral
Now, substituting this back into the integral:
\[ I = 2 \int \frac{1}{5\left(\left(t - \frac{4}{5}\right)^2 + \frac{9}{25}\right)} \, dt \]
This can be simplified to:
\[ I = \frac{2}{5} \int \frac{1}{\left(t - \frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2} \, dt \]

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

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

### Final Answer
Thus, the final answer is:
\[ \int \frac{1}{5 - 4 \sin x} \, dx = \frac{2}{3} \tan^{-1} \left( \frac{5 \tan \frac{x}{2} - 4}{3} \right) + C \]