To solve the integral $\int \sqrt{x^2 - a^2} \, dx$: Step 1: Identify the form of the integral. This integral is of the form $\int \sqrt{x^2 - a^2} \, dx$, which suggests using a trigonometric substitution. Step 2: Choose the appropriate trigonometric substitution. For expressions involving $\sqrt{x^2 - a^2}$, the standard substitution is $x = a \sec \theta$. From this substitution, we can find $dx$: $$ dx = a \sec \theta \tan \theta \, d\theta $$ Also, we can express $\sqrt{x^2 - a^2}$ in terms of $\theta$: $$ \sqrt{x^2 - a^2} = \sqrt{(a \sec \theta)^2 - a^2} = \sqrt{a^2 \sec^2 \theta - a^2} $$ $$ = \sqrt{a^2 (\sec^2 \theta - 1)} = \sqrt{a^2 \tan^2 \theta} = a |\tan \theta| $$ Assuming $x > a$ (or $x < -a$) and $\theta$ is in an appropriate range where $\tan \theta \ge 0$, we have $\sqrt{x^2 - a^2} = a \tan \theta$. Step 3: Substitute into the integral. $$ \int (a \tan \theta) (a \sec \theta \tan \theta) \, d\theta $$ $$ \int a^2 \tan^2 \theta \sec \theta \, d\theta $$ Step 4: Rewrite $\tan^2 \theta$ using the identity $\tan^2 \theta = \sec^2 \theta - 1$. $$ \int a^2 (\sec^2 \theta - 1) \sec \theta \, d\theta $$ $$ a^2 \int (\sec^3 \theta - \sec \theta) \, d\theta $$ $$ a^2 \left( \int \sec^3 \theta \, d\theta - \int \sec \theta \, d\theta \right) $$ Step 5: Evaluate the standard integrals. We use the known integral formulas: $$ \int \sec \theta \, d\theta = \ln|\sec \theta + \tan \theta| + C_1 $$ $$ \int \sec^3 \theta \, d\theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln|\sec \theta + \tan \theta| + C_2 $$ Substitute these back into the expression: $$ a^2 \left( \left( \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln|\sec \theta + \tan \theta| \right) - \ln|\sec \theta + \tan \theta| \right) + C $$ $$ a^2 \left( \frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln|\sec \theta + \tan \theta| \right) + C $$ $$ \frac{a^2}{2} (\sec \theta \tan \theta - \ln|\sec \theta + \tan \theta|) + C $$ Step 6: Convert back to $x$. From $x = a \sec \theta$, we have $\sec \theta = \frac{x}{a}$. To find $\tan \theta$, we use the identity $\tan \theta = \sqrt{\sec^2 \theta - 1}$: $$ \tan \theta = \sqrt{\left(\frac{x}{a}\right)^2 - 1} = \sqrt{\frac{x^2}{a^2} - 1} = \sqrt{\frac{x^2 - a^2}{a^2}} = \frac{\sqrt{x^2 - a^2}}{a} $$ Now substitute these expressions back into the result: $$ \frac{a^2}{2} \left( \frac{x}{a} \cdot \frac{\sqrt{x^2 - a^2}}{a} - \ln\left|\frac{x}{a} + \frac{\sqrt{x^2 - a^2}}{a}\right| \right) + C $$ $$ \frac{a^2}{2} \left( \frac{x \sqrt{x^2 - a^2}}{a^2} - \ln\left|\frac{x + \sqrt{x^2 - a^2}}{a}\right| \right) + C $$ Distribute $\frac{a^2}{2}$: $$ \frac{x \sqrt{x^2 - a^2}}{2} - \frac{a^2}{2} \ln\left|\frac{x + \sqrt{x^2 - a^2}}{a}\right| + C $$ Using the logarithm property $\ln\left|\frac{A}{B}\right| = \ln|A| - \ln|B|$: $$ \frac{x \sqrt{x^2 - a^2}}{2} - \frac{a^2}{2} (\ln|x + \sqrt{x^2 - a^2}| - \ln|a|) + C $$ $$ \frac{x \sqrt{x^2 - a^2}}{2} - \frac{a^2}{2} \ln|x + \sqrt{x^2 - a^2}| + \frac{a^2}{2} \ln|a| + C $$ Since $\frac{a^2}{2} \ln|a|$ is a constant, it can be absorbed into the arbitrary constant $C$. The final answer is $\boxed{\frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln|x + \sqrt{x^2 - a^2}| + C}$. That's 2 down. 3 left today — send the next one.