Let's solve each part step-by-step: --- ### **a. Trigonometric substitution for \(\int \frac{\sqrt{x^2-1}}{x} dx\)** #### **Step 1: Let \( x = \sec\theta \)** - Then \( dx = \sec\theta\tan\theta d\theta \) - \( x^2 - 1 = \sec^2\theta - 1 = \tan^2\theta \) - \( \sqrt{x^2-1} = \tan\theta \) So, substitute into the integral: \[ \int \frac{\sqrt{x^2-1}}{x} dx = \int \frac{\tan\theta}{\sec\theta} \cdot \sec\theta\tan\theta d\theta \] \[ = \int \tan^2\theta d\theta \] #### **Step 2: Simplify \(\tan^2\theta\)** \[ \int \tan^2\theta d\theta = \int (\sec^2\theta - 1) d\theta = \int \sec^2\theta d\theta - \int d\theta = \tan\theta - \theta + C \] #### **Step 3: Convert back to \(x\)** Recall: - \( x = \sec\theta \implies \theta = \arcsec(x) \) - \( \tan\theta = \sqrt{x^2-1} \) (since \( \tan^2\theta = x^2-1 \) and \( x > 1 \)) So, \[ \boxed{ \int \frac{\sqrt{x^2-1}}{x} dx = \sqrt{x^2-1} - \arcsec(x) + C } \] --- ### **b. Reconciling the Two Answers** From the table: - \( \int \frac{\sqrt{u^2-a^2}}{u} du = \sqrt{u^2-a^2} - \arccos\left(\frac{a}{|u|}\right) + C \) - \( \int \frac{\sqrt{u^2-a^2}}{u} du = \sqrt{u^2-a^2} - \arcsec\left(\frac{|u|}{a}\right) + C \) Recall the identity for \( u > 0 \) and \( u \geq a \): \[ \arcsec\left(\frac{u}{a}\right) = \arccos\left(\frac{a}{u}\right) \] Thus, the two answers are equivalent; they just use different inverse trigonometric functions but express the same value. --- ### **c. Evaluate \(\int \sqrt{e^{2x}-6} dx\)** Let’s use the formula from (b) with \( u = e^x \), \( a = \sqrt{6} \): \[ \int \sqrt{e^{2x}-6} dx = \int \frac{\sqrt{u^2-6}}{u} du \qquad (du = e^x dx \implies dx = \frac{du}{u}) \] But we actually have: \[ \sqrt{e^{2x}-6} = \sqrt{(e^x)^2-6} \] But let's try substitution directly: Let \( u = e^x \implies du = e^x dx \implies dx = \frac{du}{u} \) So, \[ \int \sqrt{e^{2x}-6} dx = \int \sqrt{u^2-6} \cdot \frac{du}{u} = \int \frac{\sqrt{u^2-6}}{u} du \] Now use the result from part (b) with \( a = \sqrt{6} \): \[ \int \frac{\sqrt{u^2-6}}{u} du = \sqrt{u^2-6} - \arcsec\left(\frac{u}{\sqrt{6}}\right) + C \] Substitute back \( u = e^x \): \[ \boxed{ \int \sqrt{e^{2x}-6} dx = \sqrt{e^{2x}-6} - \arcsec\left(\frac{e^x}{\sqrt{6}}\right) + C } \] --- ## **Summary of Answers** 1. **(a)** \( \int \frac{\sqrt{x^2-1}}{x} dx = \sqrt{x^2-1} - \arcsec(x) + C \) 2. **(b)** The answers are equivalent because \( \arcsec(u/a) = \arccos(a/u) \) for \( u > 0 \). 3. **(c)** \( \int \sqrt{e^{2x}-6} dx = \sqrt{e^{2x}-6} - \arcsec\left(\frac{e^x}{\sqrt{6}}\right) + C \)

Let's carefully work through each part of the problem step-by-step, providing detailed explanations and calculations. --- ## **Part (a): Use trigonometric substitution to evaluate \(\displaystyle \int \frac{\sqrt{x^2 - 1}}{x} \, dx\)** ### **Step 1: Recognize the form and choose substitution** The integrand involves \(\sqrt{x^2 - 1}\), which suggests a substitution involving secant, because: \[ x = \sec \theta \quad \Rightarrow \quad \sqrt{x^2 - 1} = \sqrt{\sec^2 \theta - 1} = \sqrt{\tan^2 \theta} = |\tan \theta| \] Assuming \(x > 1\), then \(\theta \in [0, \pi/2)\), and \(\tan \theta \ge 0\). --- ### **Step 2: Set substitution** \[ x = \sec \theta \quad \Rightarrow \quad dx = \sec \theta \tan \theta \, d\theta \] \[ \sqrt{x^2 - 1} = \tan \theta \] Rewrite the integral: \[ \int \frac{\sqrt{x^2 - 1}}{x} \, dx = \int \frac{\tan \theta}{\sec \theta} \cdot \sec \theta \tan \theta \, d\theta \] Simplify numerator and denominator: \[ \frac{\tan \theta}{\sec \theta} \times \sec \theta \tan \theta = \tan \theta \times \tan \theta = \tan^2 \theta \] So, the integral reduces to: \[ \boxed{ \int \tan^2 \theta \, d\theta } \] --- ### **Step 3: Integrate \(\tan^2 \theta\)** Recall the identity: \[ \tan^2 \theta = \sec^2 \theta - 1 \] Thus, \[ \int \tan^2 \theta \, d\theta = \int (\sec^2 \theta - 1) \, d\theta = \int \sec^2 \theta \, d\theta - \int 1 \, d\theta \] We know: \[ \int \sec^2 \theta \, d\theta = \tan \theta + C \] and \[ \int 1 \, d\theta = \theta + C \] Therefore, \[ \int \tan^2 \theta \, d\theta = \tan \theta - \theta + C \] --- ### **Step 4: Back-substitute to \(x\)** Recall: \[ x = \sec \theta \quad \Rightarrow \quad \theta = \arcsec x \] \[ \tan \theta = \sqrt{x^2 - 1} \] So, the integral becomes: \[ \boxed{ \int \frac{\sqrt{x^2 - 1}}{x} \, dx = \sqrt{x^2 - 1} - \arcsec x + C } \] --- ## **Part (b): Reconciling the two integral expressions** Given in the problem: \[ \int \frac{\sqrt{u^2 - a^2}}{u} \, du = \sqrt{u^2 - a^2} - \arccos \left( \frac{a}{u} \right) + C \] and \[ \int \frac{\sqrt{u^2 - a^2}}{u} \, du = \sqrt{u^2 - a^2} - \arcsec \left( \frac{u}{a} \right) + C \] ### **How to reconcile this?** Recall the relationship between \(\arccos\) and \(\arcsec\): \[ \arcsec y = \arccos \left( \frac{1}{y} \right) \quad \text{for } y \ge 1 \] Similarly, \[ \arccos \left( \frac{a}{u} \right) = \arccos \left( \frac{1}{\frac{u}{a}} \right) = \arcsec \left( \frac{u}{a} \right) \] **Conclusion:** \[ \boxed{ \arccos \left( \frac{a}{u} \right) = \arcsec \left( \frac{u}{a} \right) } \] Thus, both expressions are equivalent, differing only by the use of \(\arccos\) versus \(\arcsec\). They are just different inverse trigonometric functions representing the same angle. --- ## **Part (c): Use either integral to evaluate \(\displaystyle \int \sqrt{e^{2x} - 6} \, dx \)** ### **Step 1: Substitution** Let: \[ u = e^x \quad \Rightarrow \quad du = e^x dx = u \, dx \quad \Rightarrow \quad dx = \frac{du}{u} \] Express the integral in terms of \(u\): \[ \int \sqrt{u^2 - 6} \, dx = \int \sqrt{u^2 - 6} \times \frac{du}{u} = \int \frac{\sqrt{u^2 - 6}}{u} \, du \] ### **Step 2: Apply the formula from part (b)** From part (b), with \( a = \sqrt{6} \): \[ \int \frac{\sqrt{u^2 - a^2}}{u} \, du = \sqrt{u^2 - a^2} - \arcsec \left( \frac{u}{a} \right) + C \] Substitute \(a = \sqrt{6}\): \[ \int \frac{\sqrt{u^2 - 6}}{u} \, du = \sqrt{u^2 - 6} - \arcsec \left( \frac{u}{\sqrt{6}} \right) + C \] ### **Step 3: Back-substitute \(u = e^x\)** \[ \boxed{ \int \sqrt{e^{2x} - 6} \, dx = \sqrt{e^{2x} - 6} - \arcsec \left( \frac{e^x}{\sqrt{6}} \right) + C } \] --- ## **Final answers:** ### **(a)** \[ \boxed{ \int \frac{\sqrt{x^2 - 1}}{x} \, dx = \sqrt{x^2 - 1} - \arcsec x + C } \] ### **(b)** The two integral expressions are equivalent because: \[ \arccos \left( \frac{a}{u} \right) = \arcsec \left( \frac{u}{a} \right) \] They differ only by the choice of inverse trig function, but represent the same angle. ### **(c)** \[ \boxed{ \int \sqrt{e^{2x} - 6} \, dx = \sqrt{e^{2x} - 6} - \arcsec \left( \frac{e^x}{\sqrt{6}} \right) + C } \] --- **Hope this detailed step-by-step explanation helps clarify each part!**