Evaluate the integral: I = integral dx/(sqrt(3x² - 2))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{3x^2 - 2}}. $$
Explanation of Method
When $x^2$ carries a coefficient inside a radical of the form $\sqrt{ax^2-b}$, the secant substitution is adapted so that the coefficient is absorbed into the value being squared: choosing $x=\sqrt{b/a}\sec\theta$ makes $ax^2-b$ collapse to a multiple of $\tan^2\theta$, exactly as in the simpler case.
- Choose a secant substitution scaled to absorb the leading coefficient.
- Simplify the square root and $dx$ in terms of $\theta$.
- Integrate the resulting constant multiple of $\sec\theta$.
- Back-substitute to express the answer in terms of $x$.
Step 1 · Use a standard trigonometric substitution
We recognize that the integrand is of the form $\displaystyle \int \frac{dx}{\sqrt{ax^2-b}}$, which suggests using the trigonometric substitution:
$$ x = \sqrt{\frac{2}{3}}\sec(\theta). $$
This substitution works because $\sec^2(\theta)=1+\tan^2(\theta)$, which will simplify the square root expression.
Step 2 · Find expressions for $dx$ and the square root
Using the substitution $x=\sqrt{\frac23}\sec(\theta)$, we differentiate to find $dx$:
$$ dx = \sqrt{\frac{2}{3}}\sec(\theta)\tan(\theta)\,d\theta. $$
Next, we simplify the square root term:
$$ 3x^2-2 = 3\left(\frac{2}{3}\sec^2(\theta)\right) - 2 = 2\sec^2(\theta)-2 = 2(\sec^2(\theta)-1) = 2\tan^2(\theta). $$
Thus:
$$ \sqrt{3x^2-2} = \sqrt{2}\tan(\theta). $$
Step 3 · Substitute into the integral
Substituting $x=\sqrt{\frac23}\sec(\theta)$, $dx=\sqrt{\frac23}\sec(\theta)\tan(\theta)\,d\theta$, and $\sqrt{3x^2-2}=\sqrt2\tan(\theta)$ into the original integral:
$$ I = \int \frac{\sqrt{\frac23}\sec(\theta)\tan(\theta)\,d\theta}{\sqrt{2}\tan(\theta)}. $$
Simplifying:
$$ I = \int \frac{\sqrt{\frac23}}{\sqrt2}\sec(\theta)\,d\theta = \sqrt{\frac13}\int \sec(\theta)\,d\theta. $$
Step 4 · Integrate
The integral of $\sec(\theta)$ is a standard result:
$$ \int \sec(\theta)\,d\theta = \ln\left|\sec(\theta)+\tan(\theta)\right| + C. $$
Thus:
$$ I = \sqrt{\frac13}\ln\left|\sec(\theta)+\tan(\theta)\right| + C. $$
Step 5 · Back-substitute
We now need to back-substitute using $x=\sqrt{\frac23}\sec(\theta)$. From this, we get:
$$ \sec(\theta) = \frac{x}{\sqrt{\frac23}} = \sqrt{\frac32}\,x, $$
and
$$ \tan(\theta) = \sqrt{\sec^2(\theta)-1} = \sqrt{\frac32 x^2 - 1}. $$
Thus, the integral becomes:
$$ I = \sqrt{\frac13}\ln\left|\sqrt{\frac32}\,x + \sqrt{\frac32 x^2 - 1}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $x = \sqrt{2/3}\sec\theta$ |
| $\sqrt{3x^2-2}$ | $\sqrt2\tan\theta$ |
| Leading coefficient | $\sqrt{1/3}$ |
Additional Notes
- The domain of $I$ requires $x^2 \geq \tfrac23$, i.e. $|x|\geq \sqrt{2/3}$, so that $3x^2-2\geq 0$.
- Unlike Problem 44, the constant $\sqrt{1/3}$ here is a genuine multiplicative factor arising before the logarithm is taken (from simplifying $\sqrt{2/3}/\sqrt2$), not from splitting a logarithm of a quotient — so it correctly stays outside the log throughout.
- The result can be checked by differentiating $\sqrt{\tfrac13}\ln\left|\sqrt{\tfrac32}x+\sqrt{\tfrac32x^2-1}\right|$ with the chain rule and confirming it returns $\dfrac{1}{\sqrt{3x^2-2}}$.
Result
The solution to the integral is: