Calculate the integral: integral from 0 to 1 arcsin x dx

Calculate the integral: $\int_0^1 \arcsin x \, dx$.

We will use integration by parts to evaluate this integral.

Step 1: Set up integration by parts

Let $u = \arcsin x$, so $u' = \frac{1}{\sqrt{1-x^2}}$; and $v' = 1$, so $v = x$.

Step 2: Apply the integration by parts formula

\[\int_0^1 \arcsin x \, dx = \left[x \arcsin x\right]_0^1 - \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx\]

Step 3: Evaluate the boundary term

\[\left[x \arcsin x\right]_0^1 = 1 \cdot \frac{\pi}{2} - 0 = \frac{\pi}{2}\]

Step 4: Evaluate $\int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx$

Let $t = 1 - x^2$, so $dt = -2x \, dx$, i.e. $x \, dx = -\frac{1}{2} dt$.

Change the limits:

When $x=0$: $t=1$. When $x=1$: $t=0$.

Transform the integral:

\[\int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int_1^0 t^{-1/2} \, dt = -\left[\sqrt{t}\right]_1^0 = -(0-1) = 1\]

Step 5: Combine the results

\[\int_0^1 \arcsin x \, dx = \frac{\pi}{2} - 1\]

Numerical verification:

Using $\frac{\pi}{2} \approx 1.5708$: $\frac{\pi}{2} - 1 \approx 0.5708$.

Alternative form:

\[\int_0^1 \arcsin x \, dx = \frac{\pi - 2}{2}\]

Verification using the antiderivative:

The antiderivative of $\arcsin x$ is $F(x) = x \arcsin x + \sqrt{1-x^2} + C$. Check: $F'(x) = \arcsin x + \frac{x}{\sqrt{1-x^2}} - \frac{x}{\sqrt{1-x^2}} = \arcsin x$. Using this:

\[\int_0^1 \arcsin x \, dx = \left[x \arcsin x + \sqrt{1-x^2}\right]_0^1 = \left(\frac{\pi}{2} + 0\right) - (0+1) = \frac{\pi}{2} - 1\]

This confirms our result!

Geometric interpretation:

The integral represents the area under the curve $y=\arcsin x$ from $x=0$ to $x=1$. At $x=0$: $y=0$; at $x=1$: $y=\frac\pi2\approx1.571$. The function is increasing and concave down on $[0,1]$, and the area is approximately $0.571$ square units.

Final answer:

\[\boxed{\int_0^1 \arcsin x \, dx = \frac{\pi}{2} - 1}\]

Or equivalently:

\[\boxed{\int_0^1 \arcsin x \, dx = \frac{\pi - 2}{2}}\]