Question

The value of \[ \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{x \sin x}{\cos^2 x} \, dx \text{ is:} \]

• A. \( \frac{1}{3} (4\pi + 1) \)
• B. \( \frac{4\pi}{3} - 2 \log \tan \frac{\pi}{12} \)
• C. \( \frac{4\pi}{3} + \log \tan \frac{5\pi}{12} \)
• D. \( \frac{4\pi}{3} - \log \tan \frac{5\pi}{3} \)

Hint:

When dealing with integrals of odd functions over symmetric intervals, remember that you can simplify the problem by evaluating the integral over just half of the interval and doubling the result.

Answer 1

The Correct Option is B

We are given the integral: \[ I = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{x \sin x}{\cos^2 x} \, dx \] To solve this, let's first simplify the integrand: \[ \frac{x \sin x}{\cos^2 x} = x \tan x \sec x \] Next, observe that the function is odd since the product of \( x \) (an odd function) and \( \tan x \sec x \) (also odd due to the symmetry of trigonometric functions) results in an odd integrand. Now, the integral of an odd function over a symmetric interval is twice the integral over the positive half of the interval: \[ I = 2 \int_0^{\frac{\pi}{3}} \frac{x \sin x}{\cos^2 x} \, dx \] This can be solved using standard integration techniques or by looking up a known result for this specific form of integral, yielding: \[ I = \frac{4\pi}{3} - 2 \log \tan \frac{\pi}{12} \] Thus, the correct answer is \( \boxed{\text{(b)}} \).