Evaluate the integral: I = integral x²arccos x dx

Evaluate the integral: $$ I = \int x^2\arccos x\,dx. $$

Step 1 · Set up integration by parts

Let:

$$ u = \arccos x \quad \Rightarrow \quad du = \frac{-dx}{\sqrt{1-x^2}}, \qquad dv = x^2\,dx \quad \Rightarrow \quad v = \frac{x^3}{3}. $$

Applying the formula:

$$ I = \frac{x^3}{3}\arccos x + \frac13\int \frac{x^3}{\sqrt{1-x^2}}\,dx. $$

Step 2 · Evaluate the leftover integral

Let $t=1-x^2$, so $dt=-2x\,dx$ and $x^2=1-t$. Then $x^3\,dx=x^2\cdot x\,dx=(1-t)\left(-\dfrac{dt}{2}\right)$:

$$ \int \frac{x^3}{\sqrt{1-x^2}}\,dx = \int \frac{(1-t)}{\sqrt t}\left(-\frac{dt}{2}\right) = -\frac12\int \left(t^{-1/2}-t^{1/2}\right) dt. $$

Step 3 · Integrate

$$ -\frac12\left(2t^{1/2} - \frac23t^{3/2}\right) = -t^{1/2} + \frac13t^{3/2}. $$

Substituting back $t=1-x^2$:

$$ \int \frac{x^3}{\sqrt{1-x^2}}\,dx = \frac13(1-x^2)^{3/2} - (1-x^2)^{1/2}. $$

Step 4 · Combine

Substituting into the expression for $I$:

$$ I = \frac{x^3}{3}\arccos x + \frac13\left[\frac13(1-x^2)^{3/2} - (1-x^2)^{1/2}\right] + C. $$

Simplifying:

$$ I = \frac{x^3}{3}\arccos x + \frac19(1-x^2)^{3/2} - \frac13(1-x^2)^{1/2} + C. $$

Summary Table

Additional Notes

• The domain requires $-1\leq x\leq1$, where $\arccos x$ is defined.
• A common sign pitfall when computing $\int t^{-1/2}\,dt-\int t^{1/2}\,dt$ (rather than the sum) is dropping the overall minus sign from the outer $-\tfrac12$ factor when distributing it across both terms — doing so flips the sign of the whole leftover integral and produces an antiderivative that doesn't differentiate back to the original integrand.
• The result can be checked by differentiating $\tfrac{x^3}{3}\arccos x+\tfrac19(1-x^2)^{3/2}-\tfrac13(1-x^2)^{1/2}$: the non-$\arccos x$ terms cancel exactly, leaving $x^2\arccos x$.