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

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Answer 1 · 2017-04-17T17:19:46

$2sqrt(3)-pi/2$

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

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.

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