Question

If `int_0^(π/2) log cos x  dx = π/2 log(1/2)`, then `int_0^(π/2) log sec dx` = ______.

Options

• `π/2 log (1/2)`

• ` 1 - π/2 log (1/2)`

• ` 1 + π/2 log (1/2)`

• `π/2 log 2`

Answer 1

If `int_0^(π/2) log cos x  dx = π/2 log(1/2)`, then `int_0^(π/2) log sec dx` = `underlinebb(π/2 log 2)`.

Explanation:

We have `int_0^(π/2) log cos x  dx = π/2 log  1/2`  ...(i)

= `int_0^(π/2) log secx  dx`

= `int_0^(π/2) log(1/(cosx))dx`

= `-int_0^(π/2) log(cosx)dx`

= `-π/2log(1/2)`  ...[From (i)]

= `π/2log2`