`int("sin"(5x)/(3))/("sin"(x)/(2))dx` is equal to (where, C is a constant of integration)

Options

A. ` 2x + sin x + sin 2x + c `
B. ` x + 2 sin x + sin 2x + c `
C. ` 2x + sin x + 2 sin 2x + c `
D. ` x + 2 sin x + 2 sin 2x + c `

Answer

To solve the integral \(\int \frac{\sin(5x/3)}{\sin(x/2)} \, dx\), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{\sin(5x/3)}{\sin(x/2)} \, dx \]

### Step 2: Multiply and divide by \(2 \cos(x/2)\)
We can multiply and divide the integrand by \(2 \cos(x/2)\):
\[ I = \int \frac{2 \sin(5x/3) \cos(x/2)}{\sin(x/2) \cdot 2 \cos(x/2)} \, dx \]
This simplifies to:
\[ I = \int \frac{2 \sin(5x/3) \cos(x/2)}{\sin(x)} \, dx \]

### Step 3: Use the identity \(2 \sin A \cos B = \sin(A + B) + \sin(A - B)\)
Using the identity \(2 \sin A \cos B = \sin(A + B) + \sin(A - B)\), we set \(A = 5x/3\) and \(B = x/2\):
\[ 2 \sin(5x/3) \cos(x/2) = \sin\left(\frac{5x}{3} + \frac{x}{2}\right) + \sin\left(\frac{5x}{3} - \frac{x}{2}\right) \]
Calculating the angles:
\[ \frac{5x}{3} + \frac{x}{2} = \frac{10x + 3x}{6} = \frac{13x}{6} \]
\[ \frac{5x}{3} - \frac{x}{2} = \frac{10x - 3x}{6} = \frac{7x}{6} \]
Thus, we have:
\[ I = \int \frac{\sin\left(\frac{13x}{6}\right) + \sin\left(\frac{7x}{6}\right)}{\sin(x)} \, dx \]

### Step 4: Split the integral
Now we can split the integral:
\[ I = \int \frac{\sin\left(\frac{13x}{6}\right)}{\sin(x)} \, dx + \int \frac{\sin\left(\frac{7x}{6}\right)}{\sin(x)} \, dx \]

### Step 5: Use known integrals
Using the known integral results, we can evaluate these integrals:
1. \(\int \frac{\sin(kx)}{\sin(x)} \, dx = kx + C\) for \(k = \frac{13}{6}\) and \(k = \frac{7}{6}\).

Thus, we can write:
\[ I = \frac{13}{6} x + \frac{7}{6} x + C \]

### Step 6: Combine the results
Combining the results gives us:
\[ I = \frac{20}{6} x + C = \frac{10}{3} x + C \]

### Final Result
Thus, the final result of the integral is:
\[ \int \frac{\sin(5x/3)}{\sin(x/2)} \, dx = \frac{10}{3} x + C \]