Question

Evaluate the integral
${\int }_0^1\frac{\log x}{\sqrt{1-x^2}}dx$

• A. $\pi \log 2$
• B. $-\pi \log 2$
• C. $2-\log 2$
• D. $-\frac{\pi }{2}\log 2$

Answer 1

The correct option is D $-\frac{\pi }{2}\log 2$

${\int }_0^1\frac{\log x}{\sqrt{1-x^2}}dx$

$x=\sin \theta =dx=\cos \theta \cdot d\theta$

$I={\int }_0^{\frac{\pi }{2}}\log \left(\sin \theta \right)d\theta$ ----(1)

$I={\int }_0^{\frac{\pi }{2}}\log \left(\cos \theta \right)d\theta$ ----(2)

$2I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{\sin 2\theta }{2}\right)d\theta$

$2I_2={\int }_0^{\frac{\pi }{2}}\log (\sin 2\theta {)}^{d\theta }-{\int }_0^{\frac{\pi }{2}}\log 2d\theta$

${\int }_0^{\frac{\pi }{2}}\log \left(\sin 2theta\right)d\theta$

Let,

$2\theta =t=\frac{1}{2}{\int }_0^{\pi }\log \left(\sin t\right)t$

$={\int }_0^{\frac{\pi }{2}}\log \left(\sin t\right)dt$

$=I$

$2I-I={\int }_0^{\frac{\pi }{2}}\log 2d\theta$

$I=\frac{-\pi }{2}\log 2$

Or ${\int }_0^{\frac{\pi }{2}}\log \left(\sin \theta \right)d\theta =\frac{\pi }{2}\log 2$