Evaluate the integral: I = integral x²sqrt(4-x²) dx

Evaluate the integral: $$ I = \int x^2\sqrt{4-x^2}\,dx. $$

Step 1 · Substitute

Let $x=2\sin\theta$, $-\tfrac\pi2\leq\theta\leq\tfrac\pi2$, so $dx=2\cos\theta\,d\theta$ and $\sqrt{4-x^2}=2\cos\theta$:

$$ I = \int4\sin^2\theta\cdot2\cos\theta\cdot2\cos\theta\,d\theta = 16\int\sin^2\theta\cos^2\theta\,d\theta = 4\int\sin^2(2\theta)\,d\theta. $$

Step 2 · Integrate

$$ 4\int\frac{1-\cos4\theta}2\,d\theta = 2\theta - \frac{\sin4\theta}2 + C. $$

Step 3 · Convert back to $x$

With $\sin\theta=x/2$, $\cos\theta=\sqrt{4-x^2}/2$, and $\sin4\theta=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)$:

$$ \sin4\theta = 4\cdot\frac x2\cdot\frac{\sqrt{4-x^2}}2\cdot\frac{4-2x^2}4 = \frac{x(4-2x^2)\sqrt{4-x^2}}4. $$

$$ I = 2\arcsin\left(\frac x2\right) - \frac{x(2-x^2)\sqrt{4-x^2}}4 + C. $$

Summary Table

Additional Notes

• The domain is $-2\leq x\leq2$.
• The result was verified by differentiation: $\tfrac{d}{dx}\left[2\arcsin(x/2)\right]=2/\sqrt{4-x^2}$, and $\tfrac{d}{dx}\left[-\tfrac{x(2-x^2)\sqrt{4-x^2}}4\right]$ combines with it over $\sqrt{4-x^2}$ to give exactly $x^2\sqrt{4-x^2}$, matching the integrand.