91Ó°ÊÓ

To perform this substitution, we also have to compute the derivative of \(\frac{dx}{d\theta}\): $$x = 2\sin\theta$$ $$\frac{d x}{d\theta} = 2 \cos\theta$$ Now we replace \(x\) with \(2\sin\theta\) in the integral and change the differential \(d x\) using our derivative computed above.

Using the substitution, we change the integral as follows: $$\int \frac{3x + 1}{\sqrt{4 - x^2}} dx = \int \frac{3(2\sin\theta) + 1}{\sqrt{4 - (2\sin\theta)^2}} (2\cos\theta) d\theta$$

First, compute the square root in the denominator. Then, simplify the rest of the expression and integrate with respect to \(\theta\): $$\int \frac{6\sin\theta + 1}{\sqrt{4 - 4\sin^2\theta}} (2\cos\theta) d\theta = \int \frac{6\sin\theta + 1}{\sqrt{4(1 - \sin^2\theta)}} (2\cos\theta) d\theta$$ Since the square root looks like a Pythagorean identity, it simplifies to: $$\int \frac{6\sin\theta + 1}{2\cos\theta} (2\cos\theta) d\theta$$ Now, cancel out the \(2\cos\theta\) terms in the numerator and denominator and integrate: $$\int (6\sin\theta + 1) d\theta = 6\int \sin\theta d\theta + \int d\theta$$ Integrating each part individually, we have: $$6(-\cos\theta) + \theta + C = -6\cos\theta + \theta + C$$

Now, we will convert \(\theta\) back into \(x\). Recall our original substitution, \(x = 2\sin\theta\). Solving for \(\theta\), we get \(\sin\theta = \frac{x}{2}\), which means \(\theta = \arcsin(\frac{x}{2})\). Using the Pythagorean identity \(\cos^2\theta = 1 - \sin^2\theta\), we have \(\cos\theta = \pm\sqrt{1 - \sin^2\theta} = \pm\sqrt{1 - (\frac{x}{2})^2} = \pm\frac{\sqrt{4-x^2}}{2}\). Since \(\theta = \arcsin(\frac{x}{2})\) is defined in \([-\frac{\pi}{2}, \frac{\pi}{2}]\), \(\cos\theta\) is always positive. Hence, $$\cos\theta = \frac{\sqrt{4-x^2}}{2}$$ Substituting \(\theta\) and \(\cos\theta\) into the result, we get: $$-6\left(\frac{\sqrt{4-x^2}}{2}\right) + \arcsin\left(\frac{x}{2}\right) + C = -3\sqrt{4-x^2} + \arcsin\left(\frac{x}{2}\right) + C$$ Thus, the integral is: $$\int \frac{3x + 1}{\sqrt{4 - x^2}} dx = -3\sqrt{4-x^2} + \arcsin\left(\frac{x}{2}\right) + C$$