Question

How do you evaluate `\int _ { \frac { 1} { 6} \pi } ^ { \frac { 1} { 3} \pi } 3\tan ^ { 2} x d x`?

Answer 1

`2sqrt(3)-pi/2`

Explanation:

`int_(pi/6)^(pi/3)3tan^2x dx`

By the trigonometric identity: `tan^2x=sec^2x-1`,

`=3int_(pi/6)^(pi/3)(sec^2x-1)dx`

By finding antiderivatives,

`=3[tan x -x]_(pi/6)^(pi/3)`

By plugging in the upper and lower limits,

#=3[sqrt(3)-pi/3-(sqrt(3)/3-pi/6)] =3((2sqrt(3))/3-pi/6) =2sqrt(3)-pi/2#

I hope that this was clear.