Question

The integral $\int^{{7\pi /3}}_{{7\pi /4}} \sqrt {tna^2 xdx}$ is equal to:

• A. $log\,2\sqrt{2}$
• B. $log\, 2$
• C. $2\, log\, 2$
• D. $log\,\sqrt{2}$

Answer 1

Answer: (D)

Let $I=\int\limits^{{7\pi /3}}_{{7\pi /4}}$$\sqrt{tan^{2}\,x\,dx}$
$I=\int\limits^{{7\pi /3}}_{{7\pi /4}}$$tan\,x\,dx=-log \,cos\, x|^{7\pi /3}_{7\pi /4}$
$=\left[log\,cos \frac{7\pi}{3}-log\,cos\frac{7\pi }{4}\right]$
$=log\, cos\frac{7\pi }{4}-log\,cos\frac{7\pi }{3}$
$=log\left[\frac{cos\frac{7\pi }{4}}{cos\frac{7\pi }{4}}\right]=log\left[\frac{cos\left(2\pi-\frac{\pi }{4}\right)}{cos\left(2\pi +\frac{\pi}{3} \right)}\right]$
$=log\left[\frac{cos\frac{\pi }{4}}{cos\frac{\pi }{3}}\right]=log\left(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\right)$
$log\left(\frac{2}{\sqrt{2}}\right)=log\sqrt{2}.$