Question

${\int }_0^{\pi /4}{\tan }^{2}x$dx equals

• A. $\pi /4$
• B. $1+\pi /4$
• C. $1-\pi /4$
• D. None of these

Answer 1

The correct option is C $1-\pi /4$

${\int }_0^{\pi /4}{\tan }^{2}xdx$
$={\int }_0^{\pi /4}({\sec }^{2}x-1)dx$
$={[\tan x]}_0^{\pi /4}-{\left[x\right]}_0^{\pi /4}$
$=1-\frac{\pi }{4}$