Question

If `int_0^(pi/2) log (cos x) "dx" = - pi/2 log 2,` then `int_0^(pi/2) log (cosec x)`dx = ?

Options

• `pi/2 - pi/2 log 2`

• `- pi/2 log 2`

• `pi/2 log 2`

• `pi/2 + pi/2 log2`

Answer 1

`pi/2 log 2`

Explanation:

`int_0^(pi/2) log (cosec x)`dx = `int_0^(pi/2) log (sec x)`dx    ....`[int_0^"a" f(x) "dx" = int_0^"a" "f"("a - x") "dx"]`

`= int_0^(pi/2) log (1/(cos x))`dx

`= int_0^(pi/2) [log 1 - log(cos x)]` dx

`= int_0^(pi/2) log(cos x)`dx

`= pi/2 log 2`