Question

$\int_{-1}^{1} \frac{\log 2 - \log(1+x)}{\sqrt{1-x^2}} dx =$

• A. $\frac{\pi}{8}\log 2$
• B. $\frac{\pi}{2}\log 2$
• C. $\frac{\pi}{4}\log 2$
• D. $2\pi\log 2$

Hint:

For definite integrals involving $\sqrt{1-x^2}$, the substitution $x=\sin\theta$ or $x=\cos\theta$ is standard. Remember the important definite integral results: $\int_0^{\pi/2} \log(\sin x) dx = \int_0^{\pi/2} \log(\cos x) dx = -\frac{\pi}{2}\log 2$.

Answer 1

The Correct Option is D