`int(1)/((sqrt(1-x^(2)))sin^(-1)x)dx`

Explanation

To solve the integral \( \int \frac{1}{\sqrt{1 - x^2} \sin^{-1}(x)} \, dx \), we will use a substitution method. Here’s the step-by-step solution:

### Step 1: Substitution
Let \( t = \sin^{-1}(x) \). Then, we know that:
\[ x = \sin(t) \]
Differentiating both sides with respect to \( x \):
\[ dx = \cos(t) \, dt \]
Also, from the identity \( \cos(t) = \sqrt{1 - \sin^2(t)} = \sqrt{1 - x^2} \).

### Step 2: Rewrite the Integral
Now we can rewrite the integral in terms of \( t \):
\[ \int \frac{1}{\sqrt{1 - x^2} \sin^{-1}(x)} \, dx = \int \frac{1}{\sqrt{1 - \sin^2(t)} \, t} \cos(t) \, dt \]
Since \( \sqrt{1 - \sin^2(t)} = \cos(t) \), we can simplify the integral:
\[ = \int \frac{\cos(t)}{\cos(t) \, t} \, dt = \int \frac{1}{t} \, dt \]

### Step 3: Integrate
The integral \( \int \frac{1}{t} \, dt \) is a standard integral:
\[ = \ln |t| + C \]

### Step 4: Back Substitute
Now, we substitute back \( t = \sin^{-1}(x) \):
\[ = \ln |\sin^{-1}(x)| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1}{\sqrt{1 - x^2} \sin^{-1}(x)} \, dx = \ln |\sin^{-1}(x)| + C \]

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