To solve the integral

$\int \frac{dx}{\sqrt{4 - 9x^2}},$

we can use a trigonometric substitution. This is because the expression under the square root, $4 - 9x^2$, resembles the form $a^2 - u^2$, which is suitable for substitution with $x = \frac{a}{b} \sin \theta$.

Steps:

1. Rewrite the Expression: Notice that: $4 - 9x^2 = (2)^2 - (3x)^2.$ So we can set up the substitution with $3x = 2 \sin \theta$, or equivalently, $x = \frac{2}{3} \sin \theta$.

2. Differentiate $x = \frac{2}{3} \sin \theta$: $dx = \frac{2}{3} \cos \theta \, d\theta.$

3. Substitute into the Integral: Substituting $x = \frac{2}{3} \sin \theta$ and $dx = \frac{2}{3} \cos \theta \, d\theta$, the integral becomes: $\int \frac{\frac{2}{3} \cos \theta \, d\theta}{\sqrt{4 - 9 \left(\frac{2}{3} \sin \theta\right)^2}}.$

4. Simplify the Square Root: Substitute $x = \frac{2}{3} \sin \theta$ into the square root term: $4 - 9x^2 = 4 - 4 \sin^2 \theta = 4(1 - \sin^2 \theta) = 4 \cos^2 \theta.$ So, $\sqrt{4 - 9x^2} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta$.

5. Rewrite the Integral: Substitute into the integral: $\int \frac{\frac{2}{3} \cos \theta \, d\theta}{2 \cos \theta} = \int \frac{\frac{2}{3} \cos \theta \, d\theta}{2 \cos \theta}.$ Simplifying, we get: $\int \frac{1}{3} \, d\theta = \frac{\theta}{3} + C.$

6. Back-Substitute $\theta$: Recall that $x = \frac{2}{3} \sin \theta$, so $\sin \theta = \frac{3x}{2}$. Then $\theta = \arcsin\left(\frac{3x}{2}\right)$.

   Thus, the final answer is: $\int \frac{dx}{\sqrt{4 - 9x^2}} = \frac{1}{3} \arcsin\left(\frac{3x}{2}\right) + C.$

Would you like further clarification on any part of this process?

Related Questions

1. How do you solve integrals involving other trigonometric substitutions?
2. What if the integrand involved $\sqrt{9x^2 - 4}$ instead?
3. Can we use hyperbolic substitution for integrals of this form?
4. How would you solve integrals of the form $\int \frac{dx}{\sqrt{a^2 + x^2}}$?
5. How can you determine when trigonometric substitution is the best method?

Tip

Trigonometric substitution is especially useful when dealing with expressions under square roots that match the forms $a^2 - x^2$, $a^2 + x^2$, or $x^2 - a^2$.