Some trigonometric substitutions | Part-5

Integration of \( \sqrt{x^2 + a^2} \) Using Trigonometric Substitution

The integral of \( \sqrt{x^2 + a^2} \) often appears in arc length problems, area calculations, and physics applications. To evaluate:
\[ \int \sqrt{x^2 + a^2} \, dx \] we use trigonometric substitution.

Step 1: Use the substitution \( x = a \tan\theta \)

This transforms the radical expression:
\[ \sqrt{x^2 + a^2} = \sqrt{a^2 \tan^2\theta + a^2} = a \sec\theta \] and the differential becomes:
\[ dx = a \sec^2\theta \, d\theta \]

Step 2: Substitute into the integral

\( \int \sqrt{x^2 + a^2} \, dx = \int a \sec\theta \cdot a \sec^2\theta \, d\theta = a^2 \int \sec^3\theta \, d\theta \)

Step 3: Integrate \( \sec^3\theta \)

The standard result:
\[ \int \sec^3\theta \, d\theta =\] \[\frac{1}{2} \sec\theta \tan\theta + \frac{1}{2} \log | \sec\theta + \tan\theta | + C \] So:
\[ \int \sqrt{x^2 + a^2} \, dx =\] \[a^2 \left( \frac{1}{2} \sec\theta \tan\theta + \frac{1}{2} \log | \sec\theta + \tan\theta | \right) + C \]

Step 4: Substitute back in terms of \(x\)

From the substitution \( x = a \tan\theta \):

• \( \tan\theta = \frac{x}{a} \)
• \( \sec\theta = \frac{\sqrt{x^2 + a^2}}{a} \)

Final Result:

\( \int \sqrt{x^2 + a^2} \, dx =\) \(\frac{x \sqrt{x^2 + a^2}}{2} + \frac{a^2}{2} \log \left| x + \sqrt{x^2 + a^2} \right| + C \)