`I= int ( dx)/( ("arc" cos x)^(5) sqrt(1-x^2) )`.

Answer

To solve the integral

\[ I = \int \frac{dx}{(\cos^{-1} x)^5 \sqrt{1 - x^2}}, \]

we will follow these steps:

### Step 1: Substitution
Let \( t = \cos^{-1} x \). Then, we differentiate \( t \) with respect to \( x \):

\[ \frac{dt}{dx} = -\frac{1}{\sqrt{1 - x^2}}. \]

This implies that

\[ dx = -\sqrt{1 - x^2} \, dt. \]

### Step 2: Express \( \sqrt{1 - x^2} \) in terms of \( t \)
From the substitution \( t = \cos^{-1} x \), we know that

\[ x = \cos t. \]

Thus,

\[ \sqrt{1 - x^2} = \sqrt{1 - \cos^2 t} = \sin t. \]

### Step 3: Substitute in the integral
Now, substituting \( dx \) and \( \sqrt{1 - x^2} \) into the integral, we get:

\[ I = \int \frac{-\sin t \, dt}{t^5 \sin t}. \]

The \( \sin t \) terms cancel out:

\[ I = -\int \frac{dt}{t^5}. \]

### Step 4: Integrate
Now we can integrate:

\[ I = -\left(-\frac{1}{4t^4}\right) + C = \frac{1}{4t^4} + C. \]

### Step 5: Substitute back for \( t \)
Recall that \( t = \cos^{-1} x \). Therefore, we substitute back:

\[ I = \frac{1}{4(\cos^{-1} x)^4} + C. \]

### Final Answer
The final result of the integral is:

\[ I = \frac{1}{4(\cos^{-1} x)^4} + C. \]

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