To solve the integral:

$\int \frac{x}{\sqrt{4 - x^2}} \, dx$

we can proceed by using a substitution method. Here’s a step-by-step solution:

Step 1: Use Trigonometric Substitution

Since the expression under the square root is of the form $4 - x^2$, we can use the substitution: $x = 2\sin \theta$ which implies that $dx = 2\cos \theta \, d\theta$

Then, the expression $4 - x^2$ becomes: $4 - x^2 = 4 - 4\sin^2 \theta = 4\cos^2 \theta$ so that $\sqrt{4 - x^2} = 2\cos \theta$

Step 2: Substitute into the Integral

Now we can rewrite the integral in terms of $\theta$: $\int \frac{x}{\sqrt{4 - x^2}} \, dx = \int \frac{2\sin \theta}{2\cos \theta} \cdot 2\cos \theta \, d\theta$

Simplifying this expression: $= \int 2\sin \theta \, d\theta$

Step 3: Integrate with Respect to $\theta$

The integral now becomes: $\int 2\sin \theta \, d\theta = -2\cos \theta + C$

Step 4: Substitute Back in Terms of $x$

Since we originally set $x = 2\sin \theta$, we can solve for $\cos \theta$: $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{x}{2}\right)^2} = \frac{\sqrt{4 - x^2}}{2}$

Thus, $-2\cos \theta = -\sqrt{4 - x^2}$

Final Answer

The integral simplifies to: $\int \frac{x}{\sqrt{4 - x^2}} \, dx = -\sqrt{4 - x^2} + C$

Would you like a deeper explanation on any of these steps?

Here are some additional questions you might find useful:

1. How does trigonometric substitution help in evaluating integrals with roots?
2. What other types of integrals can benefit from trigonometric substitution?
3. Can we evaluate this integral using a different substitution method?
4. How can we apply this method to similar integrals involving different constants?
5. What are the practical applications of solving integrals of this form?

Tip: For integrals involving $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$, trigonometric substitutions like $x = a \sin \theta$, $x = a \tan \theta$, and $x = a \sec \theta$ are often useful.