Evaluate: `int cos^(3)x. e^(log sin x)dx`

Answer

To evaluate the integral \( \int \cos^3 x \cdot e^{\log(\sin x)} \, dx \), we can follow these steps:

### Step 1: Simplify the Integral
Using the property of logarithms and exponentials, we know that \( e^{\log(a)} = a \). Thus, we can rewrite the integral:
\[ e^{\log(\sin x)} = \sin x \]
So, the integral becomes:
\[ \int \cos^3 x \cdot \sin x \, dx \]

**Hint:** Remember that \( e^{\log(a)} = a \) for any positive \( a \).

### Step 2: Substitution
Let \( t = \sin x \). Then, the derivative \( dt = \cos x \, dx \), which implies \( dx = \frac{dt}{\cos x} \). We also need to express \( \cos^3 x \) in terms of \( t \):
\[ \cos^2 x = 1 - \sin^2 x = 1 - t^2 \]
Thus,
\[ \cos^3 x = \cos^2 x \cdot \cos x = (1 - t^2) \cdot \sqrt{1 - t^2} \]
However, since we need \( \cos^3 x \) and \( \cos x \, dx \) in terms of \( t \), we can express:
\[ \cos^3 x \cdot \sin x \, dx = (1 - t^2) \cdot \cos x \cdot dt \]

**Hint:** Use the substitution \( t = \sin x \) to simplify the integral.

### Step 3: Rewrite the Integral
Now, substituting \( \sin x = t \) and \( \cos x \, dx = dt \):
\[ \int \cos^3 x \cdot \sin x \, dx = \int (1 - t^2) t \, dt \]
This simplifies to:
\[ \int (t - t^3) \, dt \]

**Hint:** After substitution, rewrite the integral in terms of \( t \).

### Step 4: Integrate
Now we can integrate term by term:
\[ \int (t - t^3) \, dt = \frac{t^2}{2} - \frac{t^4}{4} + C \]

**Hint:** Remember to apply the power rule for integration.

### Step 5: Back Substitute
Now, substitute back \( t = \sin x \):
\[ \frac{\sin^2 x}{2} - \frac{\sin^4 x}{4} + C \]

**Hint:** Always substitute back to the original variable after integration.

### Final Result
Thus, the evaluated integral is:
\[ \int \cos^3 x \cdot e^{\log(\sin x)} \, dx = \frac{\sin^2 x}{2} - \frac{\sin^4 x}{4} + C \]