`int sin^(2//3) x cos^(3) x dx`

Answer

To solve the integral \( \int \sin^{\frac{2}{3}} x \cos^3 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We can express \( \cos^3 x \) as \( \cos^2 x \cdot \cos x \). Thus, we rewrite the integral as:
\[ \int \sin^{\frac{2}{3}} x \cos^2 x \cos x \, dx = \int \sin^{\frac{2}{3}} x \cos^2 x \, dx \cdot \cos x \]

### Step 2: Use the identity for \( \cos^2 x \)
Recall that \( \cos^2 x = 1 - \sin^2 x \). We substitute this into the integral:
\[ \int \sin^{\frac{2}{3}} x (1 - \sin^2 x) \cos x \, dx \]

### Step 3: Substitute \( t = \sin x \)
Let \( t = \sin x \). Then, \( dt = \cos x \, dx \). This transforms our integral into:
\[ \int t^{\frac{2}{3}} (1 - t^2) \, dt \]

### Step 4: Expand the integrand
Distributing \( t^{\frac{2}{3}} \) gives:
\[ \int (t^{\frac{2}{3}} - t^{\frac{8}{3}}) \, dt \]

### Step 5: Integrate term by term
Now we integrate each term:
\[ \int t^{\frac{2}{3}} \, dt - \int t^{\frac{8}{3}} \, dt \]
The integrals are:
\[ \frac{t^{\frac{2}{3} + 1}}{\frac{2}{3} + 1} - \frac{t^{\frac{8}{3} + 1}}{\frac{8}{3} + 1} + C \]
Calculating the exponents:
\[ \frac{t^{\frac{5}{3}}}{\frac{5}{3}} - \frac{t^{\frac{11}{3}}}{\frac{11}{3}} + C \]
This simplifies to:
\[ \frac{3}{5} t^{\frac{5}{3}} - \frac{3}{11} t^{\frac{11}{3}} + C \]

### Step 6: Substitute back \( t = \sin x \)
Now we substitute back \( t = \sin x \):
\[ \frac{3}{5} \sin^{\frac{5}{3}} x - \frac{3}{11} \sin^{\frac{11}{3}} x + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \sin^{\frac{2}{3}} x \cos^3 x \, dx = \frac{3}{5} \sin^{\frac{5}{3}} x - \frac{3}{11} \sin^{\frac{11}{3}} x + C \]