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

Question

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

UpStudy AI · Accepted Answer

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.

$$ \int \frac{\mathrm{d} x}{\sqrt{1-16 x^{2}}} = \arcsin\left(\frac{x}{4}\right) + C $$

(3)

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