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

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

• A. x*arcsin(x) + sqrt(1-x^2) + C
• B. sqrt(1-x^2) + C
• C. x*arcsin(x) + x + C
• D. x*arccos(x) + sqrt(1-x^2) + C

Answer: (A)

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) \

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) \