Question

Consider the integral $I={\int }_0^{\pi }ln\left(\sin x\right)dx$.What is ${\int }_0^{\frac{\pi }{2}}ln\left(\cos x\right)dx$ equal to?

• A. $\frac{I}{2}$
• B. $I$
• C. $2I$
• D. $4I$

Answer 1

The correct option is A $\frac{I}{2}$

$I={\int }_0^{\pi }ln\left(\sin x\right)dx$

using property,

$I={\int }_0^{\frac{\pi }{2}}\left(ln\right(\sin (2\pi -x)+ln\left(\sin x\right))dx$

we know that $\sin \left(x\right)=\sin (2\pi -x)$

$I=2{\int }_0^{\frac{\pi }{2}}ln\left(\sin x\right)dx$

$\frac{I}{2}={\int }_0^{\frac{\pi }{2}}ln\left(\sin x\right)dx$

by property,

$\frac{I}{2}={\int }_0^{\frac{\pi }{2}}ln\left(\sin \right(\frac{\pi }{2}-x\left)\right)dx$

$\frac{I}{2}={\int }_0^{\frac{\pi }{2}}ln\left(\cos x\right)dx$