Evaluate the integral: I = integral dx/(sqrt(3 - 4x²))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{3 - 4x^2}}. $$
Explanation of Method
When $x^2$ under the radical carries a coefficient, the denominator matches the general form $\sqrt{a^2-b^2x^2}$ rather than the plain $\sqrt{a^2-x^2}$. The corresponding antiderivative is $\int \frac{dx}{\sqrt{a^2-b^2x^2}} = \frac1b\arcsin\left(\frac{bx}{a}\right)+C$ — note that the constant out front is $\frac1b$, not $\frac1a$, which follows from substituting $u=bx$ into the plain arcsine formula.
- Match the denominator to the general form $\sqrt{a^2-b^2x^2}$.
- Identify $a$ and $b$.
- Apply the general arcsine formula, using $\frac1b$ as the leading coefficient.
Step 1 · Recognize the standard form
The given integral is of the form:
$$ \int \frac{dx}{\sqrt{a^2-b^2x^2}}, $$
where $a^2=3$ and $b^2=4$, so $a=\sqrt3$ and $b=2$. This form can be derived from the plain arcsine integral by substituting $u=bx$, which yields the standard result:
$$ \int \frac{dx}{\sqrt{a^2-b^2x^2}} = \frac{1}{b}\arcsin\left(\frac{bx}{a}\right) + C. $$
Step 2 · Apply the formula
Substitute $a=\sqrt3$ and $b=2$ into the formula:
$$ I = \frac{1}{2}\arcsin\left(\frac{2x}{\sqrt{3}}\right) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $a$, $b$ | $\sqrt3,\ 2$ |
| Leading coefficient | $\tfrac1b = \tfrac12$ |
| Antiderivative | $\tfrac12\arcsin(2x/\sqrt3)$ |
Additional Notes
- The domain of $I$ requires $|x| < \tfrac{\sqrt3}{2}$, so that $3-4x^2 > 0$.
- The general formula $\int\frac{dx}{\sqrt{a^2-b^2x^2}}=\frac1b\arcsin\left(\frac{bx}{a}\right)+C$ can be verified by direct differentiation: $\frac{d}{dx}\left[\frac1b\arcsin\left(\frac{bx}{a}\right)\right] = \frac1b\cdot\frac{b/a}{\sqrt{1-(bx/a)^2}} = \frac{1}{a\sqrt{1-(bx/a)^2}} = \frac{1}{\sqrt{a^2-b^2x^2}}$, confirming the $\tfrac1b$ coefficient (not $\tfrac1a$) is correct.
- The result can be checked by differentiating $\tfrac12\arcsin(2x/\sqrt3)$ and confirming it returns $\dfrac{1}{\sqrt{3-4x^2}}$.
Result
The solution to the integral is: