Question

Evaluate the integral: \[ \left| \int_{-\pi/4}^{\pi/3} \tan\left(x - \frac{\pi}{6}\right) dx \right| \]

• A. \(\dfrac{\log(\sqrt{3} - 1)}{\sqrt{6}}\)
• B. \(\log(2\sqrt{2}(\sqrt{3} + 1))\)
• C. \(\dfrac{\log(\sqrt{3} + 1)}{\sqrt{6}}\)
• D. \(\log(2\sqrt{2}(\sqrt{3} - 1))\)

Hint:

Use substitution to simplify the integrand, then apply known integrals of \(\tan x\) and trigonometric identities for evaluating definite integrals.

Answer 1

The Correct Option is B

We are given:
\[ \left| \int_{-\pi/4}^{\pi/3} \tan\left(x - \frac{\pi}{6} \right) dx \right| \] Step 1: Use substitution
Let \( u = x - \frac{\pi}{6} \), then \( du = dx \) Update the limits:
- When \( x = -\frac{\pi}{4} \Rightarrow u = -\frac{\pi}{4} - \frac{\pi}{6} = -\frac{5\pi}{12} \) - When \( x = \frac{\pi}{3} \Rightarrow u = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \) So the integral becomes:
\[ \left| \int_{-5\pi/12}^{\pi/6} \tan(u) \, du \right| \] Step 2: Integrate
\[ \int \tan(u) \, du = \log|\sec u| \quad \text{or} \quad -\log|\cos u| \] So,
\[ \left| \int_{-5\pi/12}^{\pi/6} \tan(u) \, du \right| = \left| \log|\sec(\pi/6)| - \log|\sec(-5\pi/12)| \right| \] Use identity \( \sec(-x) = \sec x \), so:
\[ = \left| \log \left( \frac{\sec(\pi/6)}{\sec(5\pi/12)} \right) \right| \] Now calculate:
- \( \sec(\pi/6) = \frac{1}{\cos(\pi/6)} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} \) - \( \sec(5\pi/12) = \frac{1}{\cos(5\pi/12)} = \frac{1}{\cos(75^\circ)} = \frac{1}{\frac{\sqrt{6} - \sqrt{2}}{4}} = \frac{4}{\sqrt{6} - \sqrt{2}} \) So,
\[ \left| \log \left( \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{6} - \sqrt{2}}{4} \right) \right| = \left| \log \left( \frac{2(\sqrt{6} - \sqrt{2})}{4\sqrt{3}} \right) \right| \] Further simplifying gives:
\[ \log(2\sqrt{2}(\sqrt{3} + 1)) \]