Evaluate the integral: I = integral arcsin x dx

Evaluate the integral: $$ I = \int \arcsin x\,dx. $$

Explanation of Method

Treating $\arcsin x$ as a product $\arcsin x\cdot1$ and applying integration by parts with $u=\arcsin x$, $dv=dx$ leaves a leftover integral resolved by a simple substitution.

Choose $u=\arcsin x$, $dv=dx$.
Differentiate $u$ and integrate $dv$.
Evaluate the leftover integral via substitution.

Step 1 · Set up integration by parts

Let:

$$ u = \arcsin x \quad \Rightarrow \quad du = \frac{dx}{\sqrt{1-x^2}}, \qquad dv = dx \quad \Rightarrow \quad v = x. $$

Step 2 · Apply the formula

$$ I = x\arcsin x - \int \frac{x}{\sqrt{1-x^2}}\,dx. $$

Step 3 · Evaluate the leftover integral

Let $w=1-x^2$, so $dw=-2x\,dx$:

$$ \int \frac{x}{\sqrt{1-x^2}}\,dx = -\frac12\int \frac{dw}{\sqrt w} = -\sqrt w = -\sqrt{1-x^2}. $$

Step 4 · Combine

$$ I = x\arcsin x - \left(-\sqrt{1-x^2}\right) + C = x\arcsin x + \sqrt{1-x^2} + C. $$

Summary Table

Quantity	Value
$u$	$\arcsin x$
$dv$	$dx$
Antiderivative	$x\arcsin x+\sqrt{1-x^2}$

Additional Notes

The domain requires $-1\leq x\leq1$, where $\arcsin x$ is defined.
This is the standard template for integrating any single inverse trig function: choose $u$ as the inverse function (whose derivative is algebraic) and $dv=dx$, exactly as in Problems 106, 107, and 174.
The result can be checked by differentiating $x\arcsin x+\sqrt{1-x^2}$ and confirming it returns $\arcsin x$.

Result

The solution to the integral is:

$$ I = x\arcsin x + \sqrt{1-x^2} + C. $$