(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 analyzing the integrand. The expression under the square root, $$1 - 16x^2$$, resembles the standard form of a trigonometric substitution, specifically the form $$1 - a^2x^2$$, which suggests a substitution involving the sine function.

First, I'll recall that the integral of the form $$ \int \frac{\mathrm{d} x}{\sqrt{1 - a^2x^2}} $$ can be solved using the substitution $$ x = \frac{1}{a} \sin \theta $$. This substitution simplifies the integrand to a standard form that can be easily integrated.

Applying this substitution to our integral:
Let $$ x = \frac{1}{4} \sin \theta $$, which implies $$ \mathrm{d}x = \frac{1}{4} \cos \theta \, \mathrm{d}\theta $$.

Substituting into the integral:
$$
\int \frac{\mathrm{d} x}{\sqrt{1 - 16 x^{2}}} = \int \frac{\frac{1}{4} \cos \theta \, \mathrm{d}\theta}{\sqrt{1 - 16 \left( \frac{1}{4} \sin \theta \right)^2}} = \int \frac{\frac{1}{4} \cos \theta \, \mathrm{d}\theta}{\sqrt{1 - \sin^2 \theta}} = \int \frac{\frac{1}{4} \cos \theta \, \mathrm{d}\theta}{\cos \theta} = \frac{1}{4} \int \mathrm{d}\theta = \frac{1}{4} \theta + C
$$

Now, recalling that $$ x = \frac{1}{4} \sin \theta $$, we can express $$ \theta $$ in terms of $$ x $$:
$$
\theta = \arcsin(4x)
$$
Therefore, the integral simplifies to:
$$
\frac{1}{4} \arcsin(4x) + C
$$

But wait, let's double-check this result to ensure there are no mistakes. Another approach could be to recognize that the integral resembles the standard form of the arcsine function, which is $$ \int \frac{\mathrm{d}x}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C $$. Comparing this with our integral, it seems consistent.

Alternatively, I could consider using a different substitution, such as $$ t = 4x $$, which would simplify the integral to:
$$
\int \frac{\mathrm{d}t}{\sqrt{1 - t^2}} = \arcsin(t) + C = \arcsin(4x) + C
$$
This confirms the earlier result.

Another method involves recognizing that the integral can be expressed in terms of the inverse hyperbolic sine function, but since the integrand is positive and the limits are such that the arcsine function is appropriate, the arcsine substitution seems the most straightforward.

In conclusion, after verifying through multiple methods, the integral $$ \int \frac{\mathrm{d} x}{\sqrt{1-16 x^{2}}} $$ evaluates to $$ \frac{1}{4} \arcsin(4x) + C $$, where $$ C $$ is the constant of integration.

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

(3)

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions