Find the integral of arcsin(x) dx. What is the answer?

To find the integral of arcsin(x) with respect to x, we can use integration by parts. The formula for integration by parts is:

[

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

]

For our case, we can let:

• ( u = \arcsin(x) ) (which gives ( du = \frac{1}{\sqrt{1-x^2}} , dx ))

• ( dv = dx ) (which gives ( v = x ))

Applying integration by parts gives us:

[

\int \arcsin(x) , dx = x \arcsin(x) - \int x \cdot \frac{1}{\sqrt{1-x^2}} , dx

]

Now we need to evaluate the integral ( \int \frac{x}{\sqrt{1-x^2}} , dx ). This can be solved by recognizing that:

[

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

]

Thus, we can substitute back into our expression:

[

\int \arcsin(x) \