`int(sqrt(cosx))sinxdx`

Answer

To solve the integral \( \int \sqrt{\cos x} \sin x \, dx \), we can use a substitution method. Here’s a step-by-step solution:

### Step 1: Choose a substitution
Let \( T = \cos x \). Then, the derivative of \( T \) with respect to \( x \) is:
\[ \frac{dT}{dx} = -\sin x \implies \sin x \, dx = -dT \]

### Step 2: Substitute in the integral
Now, we can rewrite the integral in terms of \( T \):
\[ \int \sqrt{\cos x} \sin x \, dx = \int \sqrt{T} (-dT) = -\int \sqrt{T} \, dT \]

### Step 3: Rewrite the integral
We can express \( \sqrt{T} \) as \( T^{1/2} \):
\[ -\int T^{1/2} \, dT \]

### Step 4: Integrate
Now, we can integrate \( T^{1/2} \):
\[ -\left( \frac{T^{3/2}}{3/2} \right) + C = -\frac{2}{3} T^{3/2} + C \]

### Step 5: Substitute back for \( T \)
Now, we substitute back \( T = \cos x \):
\[ -\frac{2}{3} (\cos x)^{3/2} + C \]

### Final Result
Thus, the integral \( \int \sqrt{\cos x} \sin x \, dx \) evaluates to:
\[ -\frac{2}{3} (\cos x)^{3/2} + C \]