Evaluate : `int"sin"^(3) x "cos"^(4) x dx`

Answer

To evaluate the integral \( \int \sin^3 x \cos^4 x \, dx \), we can use the substitution method. Here’s a step-by-step solution:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \sin^3 x \cos^4 x \, dx \]
We can rewrite \( \sin^3 x \) as \( \sin^2 x \cdot \sin x \). This allows us to express \( \sin^2 x \) in terms of \( \cos^2 x \):
\[ \sin^2 x = 1 - \cos^2 x \]
Thus, we have:
\[ \sin^3 x = \sin^2 x \cdot \sin x = (1 - \cos^2 x) \sin x \]
Now, substituting this back into the integral gives:
\[ \int (1 - \cos^2 x) \sin x \cos^4 x \, dx \]

### Step 2: Substitution
Let’s use the substitution:
\[ t = \cos x \quad \Rightarrow \quad dt = -\sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -dt \]
Now, substituting \( t \) into the integral:
\[ \int (1 - t^2) t^4 (-dt) = -\int (1 - t^2) t^4 \, dt \]
This simplifies to:
\[ -\int (t^4 - t^6) \, dt \]

### Step 3: Integrate
Now we can integrate term by term:
\[ -\left( \frac{t^5}{5} - \frac{t^7}{7} \right) + C \]
This simplifies to:
\[ -\frac{t^5}{5} + \frac{t^7}{7} + C \]

### Step 4: Substitute Back
Now we substitute back \( t = \cos x \):
\[ -\frac{\cos^5 x}{5} + \frac{\cos^7 x}{7} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \sin^3 x \cos^4 x \, dx = \frac{\cos^7 x}{7} - \frac{\cos^5 x}{5} + C \]