More Substitution Rule — Question 2

Question 2 - Solution

Because the integrand contains $\sqrt{4-x^2}$, a trigonometric substitution is natural.

Let $$x = 2\sin\theta.$$ Then $$dx = 2\cos\theta\,d\theta, \qquad \sqrt{4-x^2}=\sqrt{4-4\sin^2\theta}=2\cos\theta.$$

Substitute into the integral: $$\int \frac{x^2}{\sqrt{4-x^2}}\,dx = \int \frac{4\sin^2\theta}{2\cos\theta}\cdot 2\cos\theta\,d\theta = \int 4\sin^2\theta\,d\theta.$$

Use the identity $$\sin^2\theta=\frac{1-\cos(2\theta)}{2}.$$

Then $$\int 4\sin^2\theta\,d\theta = 4\int \frac{1-\cos(2\theta)}{2}\,d\theta = 2\int (1-\cos(2\theta))\,d\theta.$$

Integrate: $$2\left(\theta-\frac{1}{2}\sin(2\theta)\right) = 2\theta-\sin(2\theta).$$

Return to $x$ using $$\theta=\arcsin\!\left(\frac{x}{2}\right), \qquad \sin(2\theta)=2\sin\theta\cos\theta =\frac{x}{2}\sqrt{4-x^2}.$$

Thus, $$\boxed{ 2\arcsin\!\left(\frac{x}{2}\right) -\frac{x}{2}\sqrt{4-x^2} + C }$$