`int sqrt((cosx-cos^3x)/(1-cos^3x))dx` is equal to

Options

A. `(2)/(3)sin^(-1)(cos^(3//2)x)+C`
B. `(3)/(2)sin^(-1)(cos^(3//2)x)+C`
C. `(2)/(3)cos^(-1)(cos^(3//2)x)+C`
D. non of these

Explanation

To solve the integral \( \int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
We start by simplifying the expression inside the square root:
\[ \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} = \sqrt{\frac{\cos x (1 - \cos^2 x)}{1 - \cos^3 x}} = \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}} \]
Here, we used the identity \( \sin^2 x = 1 - \cos^2 x \).

### Step 2: Rewrite the integrand
We can rewrite the integrand as:
\[ \sqrt{\cos x \sin^2 x} \cdot \frac{1}{\sqrt{1 - \cos^3 x}} = \sin x \sqrt{\cos x} \cdot \frac{1}{\sqrt{1 - \cos^3 x}} \]

### Step 3: Substitution
Let \( \cos x = t \). Then, \( -\sin x \, dx = dt \) or \( dx = -\frac{dt}{\sin x} \). We also know that \( \sin^2 x = 1 - t^2 \), so \( \sin x = \sqrt{1 - t^2} \).

Substituting these into the integral gives:
\[ \int \sin x \sqrt{\cos x} \cdot \frac{1}{\sqrt{1 - \cos^3 x}} \, dx = \int \sqrt{1 - t^2} \sqrt{t} \cdot \frac{-dt}{\sqrt{1 - t^3}} \]

### Step 4: Further substitution
Let \( t^{3/2} = v \). Then, \( \frac{3}{2} t^{1/2} dt = dv \) or \( dt = \frac{2}{3} v^{-1/3} dv \).

Substituting this into the integral, we have:
\[ \int -\sqrt{1 - t^3} \cdot \sqrt{t} \cdot \frac{2}{3} v^{-1/3} dv \]

### Step 5: Solve the integral
Now we can focus on the integral:
\[ -\frac{2}{3} \int \frac{dv}{\sqrt{1 - v^2}} \]
This integral evaluates to:
\[ -\frac{2}{3} \sin^{-1}(v) + C \]

### Step 6: Back substitution
Now substituting back \( v = t^{3/2} = (\cos x)^{3/2} \):
\[ -\frac{2}{3} \sin^{-1}((\cos x)^{3/2}) + C \]

### Final Answer
Thus, the final answer is:
\[ -\frac{2}{3} \sin^{-1}(\cos^{3/2} x) + C \]