Evaluate the integral: I = integral (arcsin x)² dx

Evaluate the integral: $$ I = \int (\arcsin x)^2\,dx. $$

Step 1 · Set up integration by parts

Let:

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

Applying the formula:

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

Step 2 · Evaluate the leftover integral by parts

For $L=\displaystyle\int \frac{x\arcsin x}{\sqrt{1-x^2}}\,dx$, note that $\dfrac{x}{\sqrt{1-x^2}}\,dx$ is (up to sign) the derivative of $-\sqrt{1-x^2}$. Let:

$$ p = \arcsin x \quad \Rightarrow \quad dp = \frac{dx}{\sqrt{1-x^2}}, \qquad dq = \frac{x}{\sqrt{1-x^2}}\,dx \quad \Rightarrow \quad q = -\sqrt{1-x^2}. $$

Applying integration by parts:

$$ L = -\sqrt{1-x^2}\arcsin x + \int \sqrt{1-x^2}\cdot\frac{dx}{\sqrt{1-x^2}} = -\sqrt{1-x^2}\arcsin x + \int dx. $$

Step 3 · Finish evaluating $L$

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

Step 4 · Combine

Substituting $L$ back into the expression for $I$:

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

Simplifying:

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

Summary Table

Additional Notes

• The domain requires $-1\leq x\leq1$, where $\arcsin x$ is defined.
• The leftover integral $L$ cannot be resolved by a plain substitution, since the integrand mixes $\arcsin x$ with an algebraic factor that isn't purely its own derivative — it genuinely requires a second, independent application of integration by parts, choosing $q=-\sqrt{1-x^2}$ as the piece that absorbs the $x/\sqrt{1-x^2}$ factor.
• The result can be checked by differentiating $x(\arcsin x)^2-2x+2\sqrt{1-x^2}\arcsin x$: this gives $(\arcsin x)^2+2x\arcsin x/\sqrt{1-x^2}-2-2x\arcsin x/\sqrt{1-x^2}+2=(\arcsin x)^2$, matching the integrand.