Question

${\int }_{-\pi /3}^{\pi /3}\frac{x\sin x}{{\cos }^{2}x}dx$ 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. $3\left[\frac{\pi }{2}\log \sin \frac{\pi }{12}\right]$
• D. none of these

Answer 1

Answer: (B)

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

${\int }_{-\pi /3}^{\pi /3}\frac{x\sin x}{{\cos }^{2}x}dx=2{\int }_0^{\pi /3}x\left(\sec x\tan x\right)dx$ $[\because$ function is even$]$

$=2[{|x.\sec x|}_0^{\pi /3}-{\int }_0^{\pi /3}\sec xdx]$

$=2[\frac{\pi }{3}.\left(2\right)-{\left|\log (\sec x+\tan x)\right|}_0^{\pi /3}]$

$=2[\frac{2\pi }{3}-\log (2+\sqrt{3})+0]$

$=2[\frac{2\pi }{3}-\log \tan \frac{5\pi }{12}]$ $[\because \tan \frac{5\pi }{12}=\tan {75}^0=2+\sqrt{3}]$