If `intsqrt((cos x-cos^(3)x)/(1-cos^(3)x))dx=f(x)+c`, then `f(x)=`

Options

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

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

First, we rewrite the integrand:

\[ \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} = \sqrt{\frac{\cos x(1 - \cos^2 x)}{1 - \cos^3 x}}. \]

Using the identity \(1 - \cos^2 x = \sin^2 x\), we can rewrite this as:

\[ \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}}. \]

### Step 2: Rewrite the Denominator

Next, we can factor the denominator \(1 - \cos^3 x\):

\[ 1 - \cos^3 x = (1 - \cos x)(1 + \cos x + \cos^2 x). \]

Thus, we can express the integrand as:

\[ \sqrt{\frac{\cos x \sin^2 x}{(1 - \cos x)(1 + \cos x + \cos^2 x)}}. \]

### Step 3: Use a Substitution

Let \(t = \cos^{3/2} x\). Then, we differentiate:

\[ dt = \frac{3}{2} \cos^{1/2} x (-\sin x) \, dx \implies dx = -\frac{2}{3} \frac{dt}{\cos^{1/2} x \sin x}. \]

Now, substituting into the integral:

\[ \int \sqrt{\frac{t^2 - t^3}{1 - t^2}} \left(-\frac{2}{3} \frac{dt}{\cos^{1/2} x \sin x}\right). \]

### Step 4: Simplify the Integral

Substituting the expressions for \(\cos^{3/2} x\) and \(\sin x\) in terms of \(t\), we can simplify the integral further.

### Step 5: Evaluate the Integral

The integral can be evaluated as:

\[ -\frac{2}{3} \int \frac{1}{\sqrt{1 - t^2}} \, dt. \]

This integral is a standard form, which gives:

\[ -\frac{2}{3} \sin^{-1}(t) + C. \]

### Step 6: Substitute Back

Finally, substituting back \(t = \cos^{3/2} x\):

\[ f(x) = -\frac{2}{3} \sin^{-1}(\cos^{3/2} x) + C. \]

Thus, the final answer is:

\[ f(x) = -\frac{2}{3} \sin^{-1}(\cos^{3/2} x). \]

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