`int sin^(5//2)x cos^(3)x dx = 2 sin^(A//2)x[(1)/(B)-(1)/(C) sin^(2)x]+D`, then the value of `(A+B)-C` is equal to ........ .

Answer

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

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \sin^{5/2} x \cos^3 x \, dx \]
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can express \( \cos^3 x \) as:
\[ \cos^3 x = \cos^2 x \cdot \cos x = (1 - \sin^2 x) \cos x \]
Thus, we can rewrite the integral as:
\[ I = \int \sin^{5/2} x (1 - \sin^2 x) \cos x \, dx \]

### Step 2: Substitute \( \sin x = t \)
Let \( \sin x = t \). Then, \( \cos x \, dx = dt \). The limits of integration do not change since we are dealing with indefinite integrals. The integral becomes:
\[ I = \int t^{5/2} (1 - t^2) \, dt \]

### Step 3: Expand the integral
Now, we can expand the integral:
\[ I = \int (t^{5/2} - t^{7/2}) \, dt \]

### Step 4: Integrate term by term
Using the power rule for integration, we have:
\[ \int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C \]
Applying this to our integral:
\[ I = \left[ \frac{t^{7/2}}{7/2} - \frac{t^{9/2}}{9/2} \right] + C \]
This simplifies to:
\[ I = \frac{2}{7} t^{7/2} - \frac{2}{9} t^{9/2} + C \]

### Step 5: Substitute back \( t = \sin x \)
Now we substitute back \( t = \sin x \):
\[ I = \frac{2}{7} \sin^{7/2} x - \frac{2}{9} \sin^{9/2} x + C \]

### Step 6: Factor out common terms
Factoring out \( 2 \sin^{7/2} x \):
\[ I = 2 \sin^{7/2} x \left( \frac{1}{7} - \frac{2}{9} \sin^2 x \right) + C \]

### Step 7: Identify constants \( A, B, C \)
From the expression \( 2 \sin^{A/2} x \left( \frac{1}{B} - \frac{1}{C} \sin^2 x \right) + D \), we can identify:
- \( A = 7 \)
- \( B = 7 \)
- \( C = 9 \)

### Step 8: Calculate \( (A + B) - C \)
Now we compute:
\[ (A + B) - C = (7 + 7) - 9 = 14 - 9 = 5 \]

Thus, the final answer is:
\[ \boxed{5} \]