The value of `(pi^(2))/("In"3)int_(7//6)^(5//6) sec(pix)dx` is

Options

A. `(2pi)/(3)`
B. `(pi)/(2)`
C. `pi`
D. `(4pi)/(3)`

Answer

C

Explanation

`(pi^(2))/("In"3)overset(5//6)underset(7//6)int sec(pix)dx` `=(pi^(2))/("In"3)xx(1)/(pi)[log(sec pi x+tan pi x]_(7//6)^(5//6)`sub nothing
`=(pi)/("In"3)[log(sec(5pi)/(6)+tan(5pi)/(6))-log(("sec"7pi)/(6)+("tan"7pi)/(6))]`
`=(pi)/("In"3)[log((2)/(sqrt(3))+(1)/(sqrt(3)))-log((2)/(sqrt(3))-(1)/(sqrt(3)))]`
`=(pi)/("In"3)[logsqrt(3)-log((1)/(sqrt(3)))](pi)/("In"3)(log 3)=pi`