Evaluate the integral: I = integral dx/(sqrt(2 - 3x²))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{2 - 3x^2}}. $$
Explanation of Method
As in Problem 47, the denominator matches the general form $\sqrt{a^2-b^2x^2}$, so the antiderivative is $\frac1b\arcsin\left(\frac{bx}{a}\right)+C$, with the coefficient $\frac1b$ (not $\frac1a$) out front.
- 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 resembles the standard form:
$$ \int \frac{dx}{\sqrt{a^2-b^2x^2}}, $$
where $a^2=2$ and $b^2=3$, so $a=\sqrt2$ and $b=\sqrt3$. The standard result for this form is:
$$ \int \frac{dx}{\sqrt{a^2-b^2x^2}} = \frac{1}{b}\arcsin\left(\frac{bx}{a}\right) + C. $$
Step 2 · Apply the formula
From the standard result, substitute $a=\sqrt2$ and $b=\sqrt3$ into the formula:
$$ I = \frac{1}{\sqrt{3}}\arcsin\left(\frac{\sqrt{3}\,x}{\sqrt{2}}\right) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $a$, $b$ | $\sqrt2,\ \sqrt3$ |
| Leading coefficient | $\tfrac1b = \tfrac{1}{\sqrt3}$ |
| Antiderivative | $\tfrac{1}{\sqrt3}\arcsin(\sqrt3x/\sqrt2)$ |
Additional Notes
- The domain of $I$ requires $|x| < \sqrt{2/3}$, so that $2-3x^2 > 0$.
- Compare with Problem 47: both use the same general formula $\frac1b\arcsin\left(\frac{bx}{a}\right)+C$; here $b=\sqrt3$ is irrational, while there $b=2$ was an integer, but the formula applies identically in both cases.
- The result can be checked by differentiating $\tfrac{1}{\sqrt3}\arcsin(\sqrt3x/\sqrt2)$ and confirming it returns $\dfrac{1}{\sqrt{2-3x^2}}$.
Result
The solution to the integral is: