Some trigonometric substitutions | Part-6

Integration of \( \sqrt{x^2 - a^2} \)

The integral of the square root expression \( \sqrt{x^2 - a^2} \) often arises in problems involving hyperbolic geometry or arc lengths where the integrand includes differences of squares.

To evaluate: \[ \int \sqrt{x^2 - a^2} \, dx \] we use the substitution \( x = a \sec\theta \).
Then: \[ dx = a \sec\theta \tan\theta \, d\theta \quad\] and \[\quad \sqrt{x^2 - a^2} \] \[=\sqrt{a^2 \sec^2\theta - a^2} = a \tan\theta \]

Now substituting into the integral: \[ \int \sqrt{x^2 - a^2} \, dx = \int a \tan\theta \, a \sec\theta \tan\theta \, d\theta \] \[= a^2 \int \tan^2\theta \sec\theta \, d\theta \]

To simplify \( \int \tan^2\theta \sec\theta \, d\theta \), we rewrite: \[ \tan^2\theta = \sec^2\theta - 1 \Rightarrow \int (\sec^3\theta - \sec\theta) \, d\theta \] Using known integral formulas:

Step: Evaluate the integral

We simplify: \[ \int \tan^2\theta \sec\theta \, d\theta = \int (\sec^3\theta - \sec\theta) \, d\theta \]

Now, we apply the standard integral formulas:

Integral of \( \sec^3\theta \):
\(\int \sec^3\theta \, d\theta =\) \(\frac{1}{2} \sec\theta \tan\theta + \frac{1}{2} \log \left| \sec\theta + \tan\theta \right| + C\)
Integral of \( \sec\theta \):
\[ \int \sec\theta \, d\theta\] \[= \log \left| \sec\theta + \tan\theta \right| + C \]

Substituting these into our expression: \[ a^2 \left( \int \sec^3\theta \, d\theta - \int \sec\theta \, d\theta \right) \]
becomes:

Simplifying the logarithmic terms: \( \frac{1}{2} \log \left| \sec\theta + \tan\theta \right| - \log \left| \sec\theta + \tan\theta \right| = -\frac{1}{2} \log \left| \sec\theta + \tan\theta \right| \)

Final expression in terms of ( theta ): \( \frac{a^2}{2} \sec\theta \tan\theta - \frac{a^2}{2} \log \left| \sec\theta + \tan\theta \right| \)

Putting it all together:

Now revert back to \(x\) using \( x = a \sec\theta \), hence: \[ \sec\theta = \frac{x}{a}, \quad \tan\theta = \frac{\sqrt{x^2 - a^2}}{a} \] So, \[ \sec\theta \tan\theta = \frac{x \sqrt{x^2 - a^2}}{a^2}, \quad \] \[\sec\theta + \tan\theta = \frac{x + \sqrt{x^2 - a^2}}{a} \]

Substituting these back into the expression gives: \[ \frac{a^2}{2} \cdot \frac{x \sqrt{x^2 - a^2}}{a^2} - \frac{a^2}{2} \log \left| \frac{x + \sqrt{x^2 - a^2}}{a} \right|\] \[ = \frac{x \sqrt{x^2 - a^2}}{2} - \frac{a^2}{2} \log \left| \frac{x + \sqrt{x^2 - a^2}}{a} \right| \] We can simplify further using:

Therefore:

\[ \int \sqrt{x^2 - a^2} \, dx \] \[ =\frac{x \sqrt{x^2 - a^2}}{2} + \frac{a^2}{2} \log \left| x + \sqrt{x^2 - a^2} \right| + c \]