The integral shown in the image is:

$\int \sqrt{1 - r^2} \, dr$

This integral can be solved using a trigonometric substitution. Let's solve it step by step:

Step 1: Substitution

Let: $r = \sin \theta \quad \text{so that} \quad dr = \cos \theta \, d\theta$

Substituting $r = \sin \theta$ into the integral, we get: $\sqrt{1 - r^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta$

The integral becomes: $\int \sqrt{1 - r^2} \, dr = \int \cos \theta \cdot \cos \theta \, d\theta = \int \cos^2 \theta \, d\theta$

Step 2: Simplify $\cos^2 \theta$

Using the double-angle identity: $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$

The integral becomes: $\int \cos^2 \theta \, d\theta = \int \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int 1 \, d\theta + \frac{1}{2} \int \cos(2\theta) \, d\theta$

Step 3: Evaluate the integrals

1. For $\int 1 \, d\theta$: $\int 1 \, d\theta = \theta$

2. For $\int \cos(2\theta) \, d\theta$: $\int \cos(2\theta) \, d\theta = \frac{\sin(2\theta)}{2}$

Thus, the integral becomes: $\int \cos^2 \theta \, d\theta = \frac{1}{2} \theta + \frac{1}{4} \sin(2\theta) + C$

Step 4: Back-substitute $\theta$

Recall that $r = \sin \theta$, so: $\theta = \arcsin(r)$

Also, using the double-angle identity: $\sin(2\theta) = 2\sin\theta\cos\theta = 2r\sqrt{1 - r^2}$

Substituting back, we get: $\frac{1}{2} \theta + \frac{1}{4} \sin(2\theta) + C = \frac{1}{2} \arcsin(r) + \frac{1}{4} \cdot 2r\sqrt{1 - r^2} + C$

Simplify: $\int \sqrt{1 - r^2} \, dr = \frac{1}{2} \arcsin(r) + \frac{r\sqrt{1 - r^2}}{2} + C$

Final Answer:

$\int \sqrt{1 - r^2} \, dr = \frac{1}{2} \arcsin(r) + \frac{r\sqrt{1 - r^2}}{2} + C$

Would you like a detailed explanation of any specific step or have any other questions?

Related Questions:

1. How does trigonometric substitution work in integrals?
2. What is the purpose of the double-angle identity in integrals?
3. Can this result be used in applications like geometry or physics? If so, how?
4. How would this integral change if limits of integration were specified?
5. Are there other substitution methods to solve this integral?

Tip:

Always verify your substitution by differentiating the final result—it ensures correctness!