The integral shown is:

$\int \frac{1}{\sqrt{4 - x^2}} \, dx$

This is a standard integral that can be solved using a trigonometric substitution. The integral resembles the form of the arcsine function.

Solution:

We can rewrite $4 - x^2$ as $(2^2 - x^2)$. This suggests using the substitution:

$x = 2 \sin \theta$

From this substitution:

• $dx = 2 \cos \theta \, d\theta$
• $\sqrt{4 - x^2} = \sqrt{4 - 4 \sin^2 \theta} = 2 \cos \theta$

Now substitute these into the original integral:

$\int \frac{1}{\sqrt{4 - x^2}} \, dx = \int \frac{1}{2 \cos \theta} \cdot 2 \cos \theta \, d\theta$

This simplifies to:

$\int d\theta = \theta + C$

Since $x = 2 \sin \theta$, we have:

$\theta = \arcsin\left(\frac{x}{2}\right)$

Thus, the solution is:

$\int \frac{1}{\sqrt{4 - x^2}} \, dx = \arcsin\left(\frac{x}{2}\right) + C$

Final Answer:

$\arcsin\left(\frac{x}{2}\right) + C$

Would you like more details or have any questions?

Here are 5 follow-up questions that might interest you:

1. What are some other trigonometric substitutions used in integrals?
2. How do we handle integrals with a different constant in place of 4?
3. What is the general form of an integral that leads to arcsine?
4. Can we apply hyperbolic substitution for this type of integral?
5. How do you differentiate the arcsine function?

Tip: Always check for algebraic manipulation or trigonometric identities that simplify integral problems.