Question

\[\int\sqrt{x^2 - a^2} \text{ dx}\]

Answer 1

\[\text{ Let  I } = \int {1_{II} \cdot} \sqrt{x^2 {  _I} - a ^2} \text{ dx }\]
\[ = \sqrt{x^2 - a^2}\int1 \text{ dx} - \int\left( \frac{d}{dx}\left( \sqrt{x^2 - a^2} \right)\int1 \text{ dx}\right)dx\]
\[ = \sqrt{x^2 - a^2} \cdot x - \int\frac{1 \times 2x}{2 \sqrt{x^2 - a^2}} \cdot x\text{ dx}\]
\[ = \sqrt{x^2 - a^2} \cdot x - \int\left( \frac{x^2 - a^2 + a^2}{\sqrt{x^2 - a^2}} \right)dx\]
\[ = \sqrt{x^2 - a^2} \cdot x - \int\sqrt{x^2 - a^2} dx - a^2 \int\frac{dx}{\sqrt{x^2 - a^2}}\]
\[ = x\sqrt{x^2 - a^2} - I - a^2 \int\frac{dx}{\sqrt{x^2 - a^2}}\]
\[ \therefore 2I = x\sqrt{x^2 - a^2} - a^2 \text{ ln } \left| x + \sqrt{x^2 - a^2} \right|\]
\[ \Rightarrow I = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \text{ ln
}\left| x + \sqrt{x^2 - a^2} \right| + C\]