Question

plz help me guys......​

Answer 1

To find:-

$\displaystyle\longrightarrow\sf{\int\limits_0^{\frac {\pi}{6}}\sec^2x\ dx}$

We know that,

$\displaystyle\longrightarrow\sf{\int\sec^2x\ dx=\tan x+c}$

Thus,

$\displaystyle\longrightarrow\sf{\int\limits_0^{\frac {\pi}{6}}\sec^2x\ dx=\big [\tan x\big]_0^{\frac {\pi}{6}}}$

$\displaystyle\longrightarrow\sf{\int\limits_0^{\frac {\pi}{6}}\sec^2x\ dx=\tan\left (\dfrac {\pi}{6}\right)-\tan (0)}$

$\displaystyle\longrightarrow\sf{\int\limits_0^{\frac {\pi}{6}}\sec^2x\ dx=\dfrac {1}{\sqrt3}-0}$

$\displaystyle\longrightarrow\sf{\underline {\underline {\int\limits_0^{\frac {\pi}{6}}\sec^2x\ dx=\dfrac {1}{\sqrt3}}}}$

Hence $\displaystyle\sf {\dfrac {1}{\sqrt3}}$ is the answer.