It looks like you're asking for the integral:

\[
\int x \ln(a) \sin^{-1}(x) \, dx
\]

To solve this integral, we will use integration by parts. The integration by parts formula is given by:

\[
\int u \, dv = uv - \int v \, du
\]

Here, we can choose:

\[
u = \sin^{-1}(x), \quad dv = x \ln(a) \, dx
\]

### Step 1: Find \(du\) and \(v\)

For \(u = \sin^{-1}(x)\), the derivative \(du\) is:

\[
du = \frac{1}{\sqrt{1 - x^2}} \, dx
\]

For \(dv = x \ln(a) \, dx\), we integrate to find \(v\):

\[
v = \int x \ln(a) \, dx
\]

Since \(\ln(a)\) is a constant, we can factor it out:

\[
v = \ln(a) \int x \, dx = \ln(a) \left( \frac{x^2}{2} \right) = \frac{x^2 \ln(a)}{2}
\]

### Step 2: Apply Integration by Parts

Now, applying integration by parts:

\[
\int x \ln(a) \sin^{-1}(x) \, dx = \sin^{-1}(x) \cdot \frac{x^2 \ln(a)}{2} - \int \frac{x^2 \ln(a)}{2} \cdot \frac{1}{\sqrt{1 - x^2}} \, dx
\]

This simplifies to:

\[
\frac{x^2 \ln(a) \sin^{-1}(x)}{2} - \frac{\ln(a)}{2} \int \frac{x^2}{\sqrt{1 - x^2}} \, dx
\]

### Step 3: Simplify the Remaining Integral

To solve the integral:

\[
\int \frac{x^2}{\sqrt{1 - x^2}} \, dx
\]

We can use a substitution. Let:

\[
u = 1 - x^2 \implies du = -2x \, dx
\]

Thus, the integral becomes:

\[
-\frac{1}{2} \int \frac{1 - u}{\sqrt{u}} \, du = -\frac{1}{2} \int \left( \sqrt{u} - \frac{1}{\sqrt{u}} \right) \, du
\]

This can be integrated as:

\[
-\frac{1}{2} \left( \frac{2}{3} u^{3/2} - 2 u^{1/2} \right) = -\frac{1}{3} u^{3/2} + u^{1/2}
\]

Substituting back \(u = 1 - x^2\):

\[
-\frac{1}{3} (1 - x^2)^{3/2} + (1 - x^2)^{1/2}
\]

### Step 4: Final Answer

Substituting this back into our integration by parts formula, we get:

\[
\int x \ln(a) \sin^{-1}(x) \, dx = \frac{x^2 \ln(a) \sin^{-1}(x)}{2} - \frac{\ln(a)}{2} \left( -\frac{1}{3} (1 - x^2)^{3/2} + (1 - x^2)^{1/2} \right) + C
\]

This is the final result. Let me know if you'd like further details or if you have any questions!

### Related Questions:
1. How does integration by parts work in more complex integrals?
2. What are some alternative methods to solve integrals involving inverse trigonometric functions?
3. How do substitutions simplify integrals with square roots?
4. What are some typical mistakes to avoid when applying integration by parts?
5. How can definite integrals be solved using integration by parts?

**Tip:** When facing integrals involving inverse trigonometric functions, consider using integration by parts with the inverse function as \(u\).

Learn how to solve the integral of x ln(a) sin^(-1)(x) using integration by parts. This solution explores inverse trigonometric functions, substitution techniques, and step-by-step methods for solving advanced integrals. Suitable for college-level calculus students.

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