Question

$\int _{\pi/3}^{\pi/3} \frac {x \sin x}{\cos^2x} dx$ is equal to

• A. $\frac{\pi}{3}-\log\, \tan \frac{3\pi}{2}$
• B. $2[\frac {2\pi} {3}-\log\, \tan \frac {5\pi} {12}]$
• C. $3[\frac{\pi}{2}\log\, \sin \frac{\pi}{12}]$
• D. none of these

Answer 1

Answer: (B)

$2[\frac {2\pi} {3}-\log\, \tan \frac {5\pi} {12}]$

Answer 2

$\int\limits_{-\pi /3}^{\pi /3} \frac{x\, sin\,x}{cos^{2}\,x} dx=2$ $\int\limits_{0}^{\pi /3} x \left(sec\,x \, tan\,x\right)dx$ $=2\left[\left|x\cdot sec x\right|_{0}^{\pi /3}-\int_{0}^{\pi /3} sec\,x\,dx\right]$ $=2\left[\frac{\pi}{3}\cdot\left(2\right)-\left|log\left(sec\,x+tan\,x\right)\right|_{0}^{\pi /3}\right]$ $=2\left[\frac{2\pi}{3}-log\left(2+\sqrt{3}\right)+0\right]$ $=2\left[\frac{2\pi}{3}-log\, tan \frac{5\pi}{12}\right]$ $\left[\because tan \frac{5\pi}{12}=tan\,75^{\circ}=2+\sqrt{3}\right]$