Question

${\int }_0^{\frac{\pi }{2}}\text{log sin x dx =}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$I={\int }_0^{\frac{\pi }{2}}\text{log (sinx) dx=}{\int }_0^{\frac{\pi }{2}}\text{log (cosx) dx}\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\text{[log (sin2x)-log2] dx}\Rightarrow I=\frac{-\pi }{2}log2$