Evaluate the integral: I = integral dx/(sqrt(x² - 4))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{x^2 - 4}}. $$
Explanation of Method
The form $\sqrt{x^2-a^2}$ under a radical is a signal for the secant substitution $x=a\sec\theta$, since $\sec^2\theta-1=\tan^2\theta$ removes the square root. After substitution, the integral reduces to the classic $\int \sec\theta\,d\theta$, and back-substitution expresses the result in terms of $x$.
- Substitute $x=a\sec\theta$ to match the radical to a Pythagorean identity.
- Simplify the square root and $dx$ in terms of $\theta$.
- Integrate the resulting $\sec\theta$ using its standard antiderivative.
- 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 $\displaystyle \int \frac{dx}{\sqrt{x^2-a^2}}$, which suggests using the trigonometric substitution:
$$ x = 2\sec(\theta). $$
This is because $\sec^2(\theta)-1=\tan^2(\theta)$, and this will simplify the square root.
Step 2 · Find expressions for $dx$ and the square root
Using the substitution $x=2\sec(\theta)$, we differentiate to find $dx$:
$$ dx = 2\sec(\theta)\tan(\theta)\,d\theta. $$
Next, we simplify the square root term:
$$ \sqrt{x^2-4} = \sqrt{4\sec^2(\theta)-4} = \sqrt{4(\sec^2(\theta)-1)} = \sqrt{4\tan^2(\theta)} = 2\tan(\theta). $$
Step 3 · Substitute into the integral
Substituting $x=2\sec(\theta)$, $dx=2\sec(\theta)\tan(\theta)\,d\theta$, and $\sqrt{x^2-4}=2\tan(\theta)$ into the original integral:
$$ I = \int \frac{2\sec(\theta)\tan(\theta)\,d\theta}{2\tan(\theta)}. $$
Simplifying:
$$ I = \int \sec(\theta)\,d\theta. $$
Step 4 · Integrate
The integral of $\sec(\theta)$ is a standard result:
$$ \int \sec(\theta)\,d\theta = \ln\left|\sec(\theta)+\tan(\theta)\right| + C. $$
Step 5 · Back-substitute
We now need to back-substitute using $x=2\sec(\theta)$. From this, we get:
$$ \sec(\theta) = \frac{x}{2}. $$
Also, using the identity $\tan^2(\theta)=\sec^2(\theta)-1$, we find:
$$ \tan(\theta) = \sqrt{\sec^2(\theta)-1} = \sqrt{\left(\frac{x}{2}\right)^2-1} = \sqrt{\frac{x^2}{4}-1} = \frac{\sqrt{x^2-4}}{2}. $$
Thus, the integral becomes:
$$ I = \ln\left|\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $x = 2\sec\theta$ |
| $\sqrt{x^2-4}$ | $2\tan\theta$ |
| Reduced integral | $\int \sec\theta\,d\theta$ |
Additional Notes
- The domain of $I$ requires $|x| \geq 2$, since $x^2-4 \geq 0$ is needed for the square root to be real.
- The constant $\tfrac12$ inside the logarithm can be absorbed into $C$, so the answer is often written more simply as $\ln\left|x+\sqrt{x^2-4}\right|+C$; both forms are equivalent up to a constant shift.
- The result can be checked by differentiating $\ln\left|\frac{x+\sqrt{x^2-4}}{2}\right|$ and confirming it returns $\dfrac{1}{\sqrt{x^2-4}}$.
Result
The solution to the integral is: