`int(3sinxcos^2x-sin^3x)dx`

Answer

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

### Step 1: Rewrite the integral
We start by rewriting the integral as follows:
\[ I = \int (3\sin x \cos^2 x - \sin^3 x) \, dx \]

### Step 2: Substitute \( \cos^2 x \)
Recall that \( \cos^2 x = 1 - \sin^2 x \). We can substitute this into the integral:
\[ I = \int \left(3\sin x (1 - \sin^2 x) - \sin^3 x\right) \, dx \]

### Step 3: Expand the integrand
Now, we expand the expression inside the integral:
\[ I = \int \left(3\sin x - 3\sin^3 x - \sin^3 x\right) \, dx \]
This simplifies to:
\[ I = \int \left(3\sin x - 4\sin^3 x\right) \, dx \]

### Step 4: Separate the integral
We can separate the integral into two parts:
\[ I = \int 3\sin x \, dx - \int 4\sin^3 x \, dx \]

### Step 5: Integrate \( 3\sin x \)
The integral of \( 3\sin x \) is straightforward:
\[ \int 3\sin x \, dx = -3\cos x \]

### Step 6: Integrate \( 4\sin^3 x \)
To integrate \( 4\sin^3 x \), we can use the identity \( \sin^3 x = \sin x (1 - \cos^2 x) \):
\[ \int 4\sin^3 x \, dx = \int 4\sin x (1 - \cos^2 x) \, dx \]
This can be rewritten as:
\[ \int 4\sin x \, dx - \int 4\sin x \cos^2 x \, dx \]

The first integral is:
\[ \int 4\sin x \, dx = -4\cos x \]

For the second integral, we can use the substitution \( u = \cos x \), \( du = -\sin x \, dx \):
\[ \int 4\sin x \cos^2 x \, dx = -4 \int u^2 \, du = -4 \cdot \frac{u^3}{3} = -\frac{4\cos^3 x}{3} \]

### Step 7: Combine the results
Now, we can combine all parts:
\[ I = -3\cos x - \left(-4\cos x - \frac{4\cos^3 x}{3}\right) \]
This simplifies to:
\[ I = -3\cos x + 4\cos x + \frac{4\cos^3 x}{3} = \cos x + \frac{4\cos^3 x}{3} \]

### Step 8: Final result
Thus, the final result of the integral is:
\[ I = \cos x + \frac{4}{3}\cos^3 x + C \]
where \( C \) is the constant of integration.

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