(3) \( \int \frac{\mathrm{d} x}{\sqrt{1-16 x^{2}}} ;= \) \( \begin{array}{ll}\text { (5) } \int \frac{3 x^{3}+x}{1+x^{4}} \mathrm{~d} x & \text { (4) } \int x^{2}\left(3+2 x^{3}\right)^{\frac{1}{6}} \mathrm{~d} x\end{array} \) \( \begin{array}{ll}\text { (7) } \int \frac{\cos \ln |x|}{x} \mathrm{~d} x & \text { (6) } \int \frac{\sqrt{1+\sqrt{x}}}{\sqrt{x}} \mathrm{~d} x \\ \text { (9) } \int \frac{\cos ^{3} x}{\sin ^{2} x} \mathrm{~d} x & \text { (8) } \int \frac{\ln \ln x}{x \ln x} \mathrm{~d} x(x>e) \\ \text { (11) } \int \sin ^{2} x \cos ^{2} x \mathrm{~d} x & \text { (10) } \int \cos ^{4} x \mathrm{~d} x\end{array} \)

Alright, I need to solve the integral \( \int \frac{\mathrm{d} x}{\sqrt{1-16 x^{2}}} \). Let's start by looking at the integrand: \( \frac{1}{\sqrt{1-16 x^{2}}} \). This looks like a standard form that can be solved using a trigonometric substitution. Specifically, integrals of the form \( \int \frac{\mathrm{d} x}{\sqrt{a^2 - x^2}} \) are typically solved using the substitution \( x = a \sin \theta \). In this case, \( a = 4 \), so I'll let \( x = 4 \sin \theta \). Then, \( \mathrm{d} x = 4 \cos \theta \, \mathrm{d} \theta \). Substituting these into the integral, we get: \[ \int \frac{4 \cos \theta \, \mathrm{d} \theta}{\sqrt{1 - (4 \sin \theta)^2}} = \int \frac{4 \cos \theta \, \mathrm{d} \theta}{\sqrt{1 - 16 \sin^2 \theta}} \] Simplifying the denominator: \[ \sqrt{1 - 16 \sin^2 \theta} = \sqrt{1 - (4 \sin \theta)^2} = \sqrt{1 - 16 \sin^2 \theta} \] This doesn't seem to simplify things much. Maybe I should consider a different approach. Alternatively, I recall that integrals of the form \( \int \frac{\mathrm{d} x}{\sqrt{a^2 - x^2}} \) can also be expressed using the inverse sine function. Specifically: \[ \int \frac{\mathrm{d} x}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C \] Applying this to our integral: \[ \int \frac{\mathrm{d} x}{\sqrt{1 - 16 x^2}} = \arcsin\left(\frac{x}{4}\right) + C \] But wait, let's double-check this. If I differentiate \( \arcsin\left(\frac{x}{4}\right) \), I should get back the original integrand. The derivative of \( \arcsin(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{1 - u^2}} \), and using the chain rule: \[ \frac{\mathrm{d}}{\mathrm{d} x} \arcsin\left(\frac{x}{4}\right) = \frac{1}{\sqrt{1 - \left(\frac{x}{4}\right)^2}} \cdot \frac{1}{4} = \frac{1}{4 \sqrt{1 - \frac{x^2}{16}}} = \frac{1}{\sqrt{16 - x^2}} = \frac{1}{\sqrt{1 - 16 x^2}} \] Yes, that matches the original integrand. So, the integral evaluates to: \[ \int \frac{\mathrm{d} x}{\sqrt{1 - 16 x^2}} = \arcsin\left(\frac{x}{4}\right) + C \] But let's consider if there's another method to solve this integral to confirm. Perhaps using a substitution involving hyperbolic functions or even a substitution that simplifies the expression differently. However, the inverse sine method seems straightforward and effective here. Alternatively, I could consider completing the square or other algebraic manipulations, but they might not lead to a simpler solution. Given the standard form of the integral, the inverse sine substitution is the most efficient approach. In conclusion, the integral \( \int \frac{\mathrm{d} x}{\sqrt{1-16 x^{2}}} \) evaluates to \( \arcsin\left(\frac{x}{4}\right) + C \), where \( C \) is the constant of integration.