Question

$I_1={\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx,I_2={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\ln (\sin x+\cos x)dx$.Then

• A. $I_1=2I_2$
• B. $I_2=2I_1$
• C. $I_1=4I_2$
• D. $I_2=4I_1$

Answer 1

The correct option is A $I_1=2I_2$

$I_2={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\ln (\sin x+\cos x)dx$
$={\int }_0^{\frac{\pi }{4}}\left(\ln \right(\sin x+\cos x)+\ln (\sin (-x)+\cos (-x)\left)\right)dx$
$={\int }_0^{\frac{\pi }{4}}\left(\ln \right(\sin x+\cos x)+\ln (\cos x-\sin x\left)\right)dx$
$={\int }_0^{\frac{\pi }{4}}\ln (co{s}^{2}x-si{n}^{2}x)dx$
$={\int }_0^{\frac{\pi }{4}}\ln \left(\cos 2x\right)dx$
Putting $2x=t$,i.e.,$\frac{dt}{2}=dx$, we get
$I_2=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\cos t\right)dt=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\cos (\frac{\pi }{2}-t)\right)dt$
$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\sin t\right)dt=\frac{1}{2}{I}_1$ or $I_1=2I_2$