`int sqrt((cos x - cos^3 x)/(1 - cos^3 x)) 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. none of these

Answer

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:
\[ \frac{\cos x - \cos^3 x}{1 - \cos^3 x} \]
We can factor out \(\cos x\) from the numerator:
\[ \cos x(1 - \cos^2 x) = \cos x \sin^2 x \]
Thus, the integral becomes:
\[ \int \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}} \, dx \]

### Step 2: Rewrite the denominator
Next, we rewrite the denominator \(1 - \cos^3 x\):
\[ 1 - \cos^3 x = (1 - \cos x)(1 + \cos x + \cos^2 x) \]
This allows us to express the integral as:
\[ \int \frac{\sqrt{\cos x \sin^2 x}}{\sqrt{(1 - \cos x)(1 + \cos x + \cos^2 x)}} \, dx \]

### Step 3: Use substitution
Now, we will use the substitution:
\[ t = \cos^{3/2} x \]
Differentiating gives:
\[ dt = \frac{3}{2} \cos^{1/2} x (-\sin x) \, dx \implies dx = -\frac{2}{3} \frac{dt}{\sqrt{\cos x}} \]

### Step 4: Substitute in the integral
Substituting \(t\) into the integral, we have:
\[ \int \frac{\sin x \sqrt{\cos x}}{\sqrt{1 - t^2}} \left(-\frac{2}{3} \frac{dt}{\sqrt{\cos x}}\right) \]
This simplifies to:
\[ -\frac{2}{3} \int \frac{\sin x}{\sqrt{1 - t^2}} \, dt \]

### Step 5: Integrate
The integral \(\int \frac{\sin x}{\sqrt{1 - t^2}} \, dt\) can be recognized as:
\[ -\frac{2}{3} \sin^{-1}(t) + C \]

### Step 6: Substitute back
Now we substitute back \(t = \cos^{3/2} x\):
\[ -\frac{2}{3} \sin^{-1}(\cos^{3/2} x) + C \]

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

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