91Ó°ÊÓ

We have a sum of squares, so we choose the substitution $$x = a \sin(\theta)$$, where $$a = 3$$ in this case. This means the differential is $$d x = 3 \cos(\theta) d\theta$$. Thus, our substitution becomes $$x = 3\sin(\theta)$$.

Now we substitute for $$x$$ and $$d x$$ in the integral to get: $$\int \sqrt{4 (3\sin(\theta))^2 + 36} (3 \cos(\theta)) d\theta = 3\int \sqrt{36\sin^2(\theta) + 36} \cos(\theta) d\theta$$ Factor out the constant term from inside the square root: $$= 3 \int \sqrt{36 (\sin^2(\theta) + 1)} \cos(\theta) d\theta$$ Now, based on the Pythagorean identity, $$\sin^2(\theta) + 1 = \cos^2(\theta)$$: $$= 3 \int \sqrt{36 \cos^2(\theta)} \cos(\theta) d\theta$$ Take the square root of $$\cos^2(\theta)$$ to obtain $$\cos(\theta)$$, and the integral simplifies to: $$= 3 \int 6 \cos(\theta) \cos(\theta) d\theta$$

Integrate the simplified expression: $$= 3 \int 6 \cos^2(\theta) d\theta$$ Now, we use the double angle identity for $$\cos^2(\theta)$$: $$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$$: $$= 3 \int 6 \frac{1 + \cos(2\theta)}{2} d\theta = 9 \int \left(\frac{1}{2} + \frac{\cos(2\theta)}{2}\right) d\theta$$ Integrate each term separately: $$= 9 \left[ \frac{1}{2}\int{1\, d\theta} + \frac{1}{2}\int{\cos(2\theta) d\theta} \right]$$ $$= 9\left[\frac{1}{2}(\theta + \frac{1}{2} \frac{\sin(2\theta)}{2})\right] + C$$ where $$C$$ is the constant of integration.

Recall that $$x = 3\sin(\theta)$$. So, $$\theta = \arcsin\left(\frac{x}{3}\right)$$. Also, we need to compute $$\sin(2\theta)$$, we can use the double angle formula: $$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$. Since $$\sin(\theta) = \frac{x}{3}$$, we can use the Pythagorean identity to find $$\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - (\frac{x}{3})^2} = \frac{\sqrt{9-x^2}}{3}$$. Now we compute $$\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2\cdot\frac{x}{3}\cdot\frac{\sqrt{9-x^2}}{3} = \frac{2x\sqrt{9-x^2}}{9}$$. Plugging these values back into our result, we get: $$\int \sqrt{4 x^{2}+36} d x = 9\left[\frac{1}{2}\arcsin\left(\frac{x}{3}\right) + \frac{1}{4} \frac{2x\sqrt{9-x^2}}{9}\right] + C$$ Simplifying the expression: $$= \frac{9}{2}\arcsin\left(\frac{x}{3}\right) + \frac{1}{2}x\sqrt{9-x^2} + C$$ This is the final result for the indefinite integral of $$\sqrt{4 x^{2}+36}$$.