1. We are asked to evaluate the integral $$\int \sqrt{4 - t^2} \, dt$$. 2. This integral represents the area under the curve of the function $$y = \sqrt{4 - t^2}$$, which is the upper half of a circle with radius 2 centered at the origin. 3. To solve this integral, we use the trigonometric substitution $$t = 2\sin(\theta)$$, which implies $$dt = 2\cos(\theta) \, d\theta$$. 4. Substitute into the integral: $$\int \sqrt{4 - (2\sin(\theta))^2} \cdot 2\cos(\theta) \, d\theta = \int \sqrt{4 - 4\sin^2(\theta)} \cdot 2\cos(\theta) \, d\theta$$ 5. Simplify inside the square root: $$\sqrt{4(1 - \sin^2(\theta))} = \sqrt{4\cos^2(\theta)} = 2|\cos(\theta)|$$ 6. Since $$\theta$$ is in the range where $$\cos(\theta) \geq 0$$ (due to substitution), we have: $$2\cos(\theta)$$ 7. The integral becomes: $$\int 2\cos(\theta) \cdot 2\cos(\theta) \, d\theta = \int 4\cos^2(\theta) \, d\theta$$ 8. Use the identity: $$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$$ 9. Substitute this into the integral: $$\int 4 \cdot \frac{1 + \cos(2\theta)}{2} \, d\theta = \int 2(1 + \cos(2\theta)) \, d\theta = 2\int (1 + \cos(2\theta)) \, d\theta$$ 10. Integrate term by term: $$2\left( \theta + \frac{\sin(2\theta)}{2} \right) + C = 2\theta + \sin(2\theta) + C$$ 11. Substitute back to $$t$$: Since $$t = 2\sin(\theta)$$, then $$\sin(\theta) = \frac{t}{2}$$ and $$\theta = \arcsin\left(\frac{t}{2}\right)$$. Also, $$\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2 \cdot \frac{t}{2} \cdot \sqrt{1 - \left(\frac{t}{2}\right)^2} = t \sqrt{1 - \frac{t^2}{4}} = \frac{t \sqrt{4 - t^2}}{2}$$. 12. Therefore, the integral is: $$2 \arcsin\left(\frac{t}{2}\right) + \frac{t \sqrt{4 - t^2}}{2} + C$$ This is the exact antiderivative of the given integral.